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U.U.D.M. Project Report 2017:31
Examensarbete i matematik, 15 hpHandledare: Maksim Maydanskiy Examinator: Jörgen ÖstenssonJuni 2017
Department of MathematicsUppsala University
Hodge Decomposition for Manifolds with Boundary and Vector Calculus
Olle Eriksson
Abstract
Hodge Decomposition for Manifolds with Boundaryand Vector Calculus
Olle Eriksson
This thesis describes the Hodge decomposition of the space of differential forms on acompact Riemannian manifold with boundary, and explores how, for subdomains of3-space, it can be translated into the language of vector calculus. In the former, moregeneral, setting, we prove orthogonality of the decomposition. In the latter setting,we sketch the full proof, based on results from algebraic topology and about thesolvability of boundary value problems for certain PDEs.
Contents
1 Introduction 5
2 Differential forms 72.1 Riemannian manifolds . . . . . . . . . . . . . . . . . . . . . . 72.2 Differential forms . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Orientations and integration on manifolds . . . . . . . . . . . 102.4 An inner product on Λk(T ∗pM) . . . . . . . . . . . . . . . . . 122.5 The Hodge star operator . . . . . . . . . . . . . . . . . . . . . 142.6 An L2-inner product on differential forms . . . . . . . . . . . 182.7 The codifferential . . . . . . . . . . . . . . . . . . . . . . . . . 19
3 Vector calculus and differential forms 223.1 The Laplace-de Rham operator . . . . . . . . . . . . . . . . . 223.2 Musical isomorphisms . . . . . . . . . . . . . . . . . . . . . . 243.3 Gradient and divergence . . . . . . . . . . . . . . . . . . . . . 263.4 Euclidean space . . . . . . . . . . . . . . . . . . . . . . . . . . 273.5 Three-space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.6 Vector calculus identities . . . . . . . . . . . . . . . . . . . . . 333.7 Classical theorems . . . . . . . . . . . . . . . . . . . . . . . . 34
4 Algebraic topology 384.1 Singular homology and cohomology . . . . . . . . . . . . . . . 384.2 Poincare-Lefschetz duality . . . . . . . . . . . . . . . . . . . . 404.3 Alexander duality . . . . . . . . . . . . . . . . . . . . . . . . . 414.4 Homology in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . 424.5 De Rham cohomology . . . . . . . . . . . . . . . . . . . . . . 444.6 De Rham’s theorem . . . . . . . . . . . . . . . . . . . . . . . 45
5 Hodge theory 465.1 Notation and definitions . . . . . . . . . . . . . . . . . . . . . 465.2 The Hodge decomposition theorem . . . . . . . . . . . . . . . 505.3 Hodge isomorphism theorem . . . . . . . . . . . . . . . . . . . 525.4 Hodge decomposition in three-space . . . . . . . . . . . . . . 53
3
6 The Biot-Savart formula and boundary value problems 596.1 Dirichlet and Neumann problems . . . . . . . . . . . . . . . . 596.2 The Biot-Savart formula . . . . . . . . . . . . . . . . . . . . . 60
7 Proof of spanning statement for domains in R3 627.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 627.2 Notation and definitions . . . . . . . . . . . . . . . . . . . . . 627.3 Knots and gradients . . . . . . . . . . . . . . . . . . . . . . . 647.4 Splitting knots . . . . . . . . . . . . . . . . . . . . . . . . . . 657.5 Splitting gradients . . . . . . . . . . . . . . . . . . . . . . . . 687.6 Splitting divergence free gradients . . . . . . . . . . . . . . . 697.7 Putting everything together . . . . . . . . . . . . . . . . . . . 72
8 Bibliography 73
4
1 Introduction
Vector calculus, also known as vector analysis, is a branch of mathematicsthat extends the elements of integral and differential calculus to vectorfields defined on subsets of three-dimensional space (or some other suitablespace). Its importance is indicated by its many applications in physics andengineering, where, among other things, it is used to desribe electromagneticand gravitational fields and various flow fields. The classical operators knownas gradient, curl and divergence, typically denoted by ∇, ∇× and ∇·, andthe Laplacian, which we denote by ∇2, are important objects of study, andare related to each other by various vector calculus identities that can beused to facilitate computations.
Since a smooth three-dimensional manifold is, in a sense, nothing morethan bits and pieces of three-dimensional space that are glued togetherin a smooth and seamless way, vector calculus can be naturally extendedto this setting. But on a manifold, it is oftentimes more convenient towork with differential forms than with vector fields. In particular, on aRiemannian manifold, the Riemannian metric provides a natural (that is,basis-independent) isomorphism between vector fields and differential 1-forms.In the case of a three-dimensional Riemannian manifold, the so-called Hodgestar operator, denoted by ?, lets us extends this isomorphism to differential2-forms as well, and also lets us construct the codifferential d∗, which is,in a sense, the adjoint operator to the exterior derivative d on differentialforms. In this setting, the gradient, curl and divergence of vector analysisfind a natural generalization in the exterior derivative, and the Laplacian isgeneralized by the Laplace-de Rham operator ∆. In Chapters 2 and 3 we setup the machinery of differential forms on Riemannian manifolds and look athow the classical language of vector calculus can be translated into, and isgeneralized by, the modern language of differential geometry.
Hodge theory, named after William Vallance Douglas Hodge (1903–1975),puts the theory of partial differential equations to work to study the cohomol-ogy of smooth manifolds. That is, it studies certain topological propertiesof a manifold by means of PDEs. Central to this study is the exteriorderivative, the codifferential and the Laplace-de Rham operator. The Hodgedecomposition theorem, which lies at the heart of Hodge theory, uses theseoperators to decompose the space of differential k-forms into a direct sum of
5
L2-orthogonal subspaces. In Chapter 4 we introduce the notions of singularhomology and de Rham cohomology, and state some results that will proveto be useful later when, in Chapter 5, we state the Hodge decompositiontheorem (Theorem 5.5) as well as a special case of this theorem that appliesto vector fields on certain domains in three-space (Theorem 5.12) and thatlets us put our results from the previous chapters to the test.
Chapter 6 introduces various tools that are then used in Chapter 7 tosketch a full proof of Theorem 5.12.
Conventions
This might be a good time and place to say a few words about the conventionsused throughout this thesis. Everything in this section applies everywhere inthis thesis unless otherwise stated.
Smooth is taken to mean C∞. Manifolds are assumed to be smooth, asare differential forms and vector fields. The letter M is used to denote amanifold, and g is used to denote a Riemannian metric. The dimension ofa manifold is denoted by n. Three dimensional manifolds are sometimesdenoted by Y (this is because the letter Y resembles a three-way intersection)For domains in Rn we write D. Vector spaces are assumed to be real, andrings are assumed to be unital. The letter Γ denotes the space of smoothsections of a fiber bundle, e.g. Γ(TM) denotes the field of smooth sectionsof the tangent bundle or, in other words, the space of smooth vector fieldson M .
6
2 Differential forms
This chapter introduces the major players in what is to come: Riemannianmanifolds with boundary, differential forms, the Hodge star operator andthe codifferential. The aim is to introduce those parts of manifold theorythat are relevant to the topics that lie ahead, i.e. a discussion of the Hodgedecomposition theorem and how it can be translated into the language ofvector calculus in Euclidean 3-space.
For an introduction to smooth manifold theory and Riemannian manifolds,Lee’s books [8] and [7] are useful. A brief refresher that aims to presentthe parts of manifold theory necessary to introduce and prove the Hodgedecomposition is found in Schwarz’s book [12].
2.1 Riemannian manifolds
Let M be a smooth manifold with boundary (note that M does not haveto have a boundary). We write TM for the tangent bundle of M , and TpMfor the tangent space to M at the point p ∈M . Similarly, T ∗M denotes thectangent bundle of M , and we write T ∗pM for the cotangent space to M atp. A Riemannian metric on M is a family of positive definite inner products
gp : TpM × TpM −→ R, p ∈M
such that for all (smooth) vector fields V and W the map
p 7−→ gp(V (p),W (p))
is smooth. A Riemannian manifold (M, g) is a smooth manifold M quippedwith a Riemannian metric.
Example 2.1. The pair (Rn, ·), where Rn is understood as a smooth manifoldin the usual way and · is the dot product, is a Riemannian manifold. Wheneverwe talk about Rn as a Riemannian manifold, it is implied, unless otherwisestated, that the Riemannian metric is given by the dot product.
A useful property of Riemannian manifolds is the existence of localorthonormal frames. Given an open subset U ∈ M , a local orthonormalframe on U is a set of (not necessarily smooth) vector fields E1, . . . , En
7
defined on U that are orthonormal with respect to the Riemannian metricat each point p ∈ U , that is, gp(Ei(p), Ej(p)) = δij . It is convenient toknow that at every pont p of a Riemannian manifold there exists a localorthonormal frame on an open set containing p.
If M is oriented, the orientation of M induces an orientation of TpM foreach p ∈ M . A local orthonormal frame E1, . . . , En defined on U ∈ Mis said to be (positively) oriented if the ordered basis (E1(p), . . . , En(p)) ofTpM is an ordered basis at each p ∈ U .
Of particular interest to us is the class of smooth manifolds called regulardomains in Rn. These are properly embedded codimension 0 submanifoldswith boundary. In addition to this, we consider them as Riemannian subman-ifolds of Rn equipped with the Euclidean metric. Note that regular domainsin Rn are orientable.
2.2 Differential forms
Let M be a smooth n-manifold. We let Λk(T ∗M) denote the k-th exteriorpower of the cotangent space T ∗M . A smooth differential k-form is a smoothsection of Λk(T ∗M), i.e. a smooth map
η : M −→ Λk(T ∗M),
so that(π η)(x) = x
for all points x ∈M . Throughout this text, we will often write “differentialform”, “differential k-form”, “k-form” or just “form” when referring to asmooth section of Λk(T ∗M), omitting smooth, as smoothness of forms isalways understood. A form of degree n is sometimes called a top level form,or just a top form.
The space of k-forms on M is
Ωk(M) := Γ(Λk(T ∗M))
and the space of all differential forms on M is
Ω∗(M) :=n⊕k=0
Ωk(M).
The wedge product on k-covectors, i.e. elements of Λk(T ∗pM) at some
point p ∈M , extends pointwise to define a product on Ωk(M), also knownas the wedge product and denoted by ∧. The following proposition statessome of its properties.
Proposition 2.2. Let M be a smooth manifold. Then the following state-ments are true:
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1. The wedge product is associative, bilinear and satisfies
η ∧ ζ = (−1)klζ ∧ η
for all η ∈ Ωk(M) and ζ ∈ Ωl(M).
2. Let η1, . . . , ηk ∈ Ω1(∗M) and let v1, . . . , vk ∈ TpM for some pointp ∈M . Then
(η1 ∧ . . . ∧ ηk)p(v1, . . . , vk) = det(ηi(vj)).
The wedge product might be easiest to grasp by looking at an example.
Example 2.3. Let Y be a smooth 3-manifold and consider the differential 2-form η and 1-form ζ, which in some local coordinates are given by η = x dx∧dyand ζ = 5 dx+ y dz. Then
η ∧ ζ = (x dx ∧ dy) ∧ (5 dx+ y dz)
= 5x dx ∧ dy ∧ dx+ xy dx ∧ dy ∧ dz
= xy dx ∧ dy ∧ dz,
and since η ∧ ζ = (−1)2η ∧ ζ, we get the same result if we swap η and ζ inthe above computation.
A very important map that takes differential k-forms to (k + 1)-formsis the exterior derivative, denoted by d. It is defined to be the unique mapthat satisfies the following properties:
1. It is R-linear.
2. Let η ∈ Ωk(M) and ζ ∈ Ωl(M). Then
d(η ∧ ζ) = dη ∧ ζ + (−1)kζ ∧ dη.
3. dd = 0.
4. It is the differential for 0-forms, i.e. for smooth functions.
As with the wedge product, it is probably easiest to get a grasp of theexterior derivative by looking at an example.
Example 2.4. A 2-form η is expressible in local coordinates as η = f dx1 ∧dx2 over 1-form basis dx1, . . . , dxn. We calculate
dη =
(n∑i=1
∂f
∂xidxi
)dx1 ∧ dx2 =
(n∑i=3
∂f
∂xidxi ∧ dx1 ∧ dx2
).
9
A useful property of differential forms is that they can be pulled backfrom one smooth manifold to another via smooth maps. Let M1 and M2 besmooth manifolds with boundary, let F : M −→ N be a smooth map and letη be a k-form on N . Then the pullback of η by F is the k-form on M givenby
(F ∗η)p(V1, . . . , Vk) = ηp(dFp(V1), . . . , dFp(Vk)),
for V1, . . . .Vk ∈ TpM and for every point p ∈ M . We state the followingimportant property of the pullback and the exterior derivative without proof.
Proposition 2.5. Let M and N be smooth manifolds with boundary and letF : M −→ N be a smooth map. Then
F ∗(dη) = d(F ∗η).
As we shall see in the next chapter, the exterior derivative generalizesmany important notions from vector calculus, such as the gradient and curlof a vector field in R3.
2.3 Orientations and integration on manifolds
Let (M, g) be an oriented Riemannian manifold with boundary. An orienta-tion form on M is a nowhere vanishing top form η such that
η(E1, . . . , En) > 0
whenever E1, . . . , En is a local oriented orthonormal frame at point x ∈M .An argument using partitions of unity shows that the existence of a non-vanishing top form on M is equivalent to to M being orientable. The choiceof such a form up to multiplication by a positive function is equivalent tochoosing an orientation on M .
WheneverM is orientable, then so is ∂M . Moreover, by another partitionsof unity argument, it can be shown that there exists a smooth outwardpointing vector field N along ∂M . Let p ∈ ∂M , and let (E1, . . . , En−1)be a local oriented frame of ∂M at p. We define the orientation of Tp∂Mby declaring the orientation of (E1(p), . . . , En−1(p)) to be positive if (N(p),E1(p), . . . , En−1(p)) is positively oriented as a frame of TpM . This definesan orientation for the boundary ∂M in a consistent way.
There exists a unique orientation form called the Riemannian volumeform (or simply the volume form), on M , which we denote by ωg, and whichhas the defining property that
ωg(E1, . . . , En) = 1
for every local oriented orthonormal frame (E1, . . . , En).
10
Remark 2.6. Specifying a volume form does not determine a unique Rie-mannian metric on a smooth manifold. On the other hand, the Riemannianmetric together with an orientation is enough to determine the volume formuniquely.
We define the integral of a top form η = f dx1 ∧ . . . ∧ dxn over a domainof integration D in Rn as∫
Df dx1 ∧ . . . ∧ dxn :=
∫Df dx1 . . . dxn.
Via pullbacks to local coordinates and partitions of unity this definition canbe extended to allow for the integration of top forms over compact orientablesmooth manifolds. We will not dwell on the details here, as that is betterleft to any textbook on differential geometry, e.g. [8].
One of the most important and elegant theorems concerning the inte-gration of differential forms is Stokes’ theorem. We state it here withoutproof.
Theorem 2.7 (Stokes’ theorem). Let M be a compact orientable smoothmanifold with boundary and let η ∈ Ωk−1(M). Then∫
Mdη =
∫∂M
η.
Remark 2.8. In Stokes’ theorem, the integral over the boundary is to beinterpreted in the following way. Let ı : ∂M −→ M denote the naturalinclusion map. Then ∫
∂Mη :=
∫∂M
ı∗η,
where ı∗η denotes the pullback of η to ∂M by ı. Also, ∂M is assumed tohave the induced boundary orientation.
On an compact oriented Riemannian manifold M , the existence of thevolume form ωg lets us integrate real-valued funtions as follows. Let f ∈C∞(M). Then the integral of f over M is given by∫
Mf ωg.
In particular, by integrating f = 1 over M , the volume form lets us define“the volume” of M as
Vol(M) :=
∫Mωg.
11
2.4 An inner product on Λk(T ∗pM)
Let (M, g) be an oriented Riemannian manifold, let p ∈ M be a pointand let E1, . . . , En be an orthonormal frame at p and e1, . . . , en thecorresponding dual frame. A basis for Λk(T ∗pM) is then given by the set
B = eα | α a k-dimensional multi-index with α1 < . . . < αk.
We now define an inner product 〈·, ·〉g on Λk(T ∗pM) as follows:
〈·, ·〉g : Λk(T ∗pM)× Λk(T ∗pM)→ R
(η, ζ) 7→ 1
k!
∑1≤
i1,...,ik≤n
η(Ei1 , . . . , Eik)ζ(Ei1 , . . . , Eik).
This inner product is independent of the choice of orthonormal frame and ishence well defined. It has the property that it makes B orthonormal, whichwe will state as Proposition 2.11. But first we give an alternative definitionof 〈·, ·〉g.
Proposition 2.9. Let η, ζ ∈ Λk(T ∗pM). Then
〈η, ζ〉g =∑
σ∈S(k,n)
η(Eσ(1), . . . , Eσ(k))ζ(Eσ(1), . . . , Eσ(k)),
where S(k, n) denotes the subset of Sn consisting of all permutations σ suchthat σ(1) < . . . < σ(k) and σ(k + 1) < . . . < σ(n).
Proof. Let η, ζ ∈ Λk(T ∗pM). Then, by definition,
〈η, ζ〉g =1
k!
∑1≤
i1,...,ik≤n
η(Ei1 , . . . , Eik)ζ(Ei1 , . . . , Eik).
But since differential forms are alternating, any summand with a repeatedindex, i.e. with Eij = Eil for some j 6= l, contributes zero to the above sum.Hence∑
1≤i1,...,ik≤n
η(Ei1 , . . . , Eik)ζ(Ei1 , . . . , Eik)
=∑
1≤i1,...,ik≤nij 6=il
η(Ei1 , . . . , Eik)ζ(Ei1 , . . . , Eik)
=1
(n−k)!
∑σ∈Sn
η(Eσ(1), . . . , Eσ(k))ζ(Eσ(1), . . . , Eσ(k)). (2.1)
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Also, since each summand is the product of two alternating forms, evaluatedat the same vectors, swapping any two arguments changes the sign of bothforms, and hence the sign of a summand is unchanged under such swaps.This implies that each summand in (2.1) is invariant under permutationsof the k vectors Eσ(1), . . . , Eσ(k), and since there are k! such permutationsfor each σ ∈ Sk, of which precisely one is expressible as Eτ(1), . . . , Eτ(k) withτ ∈ S(k, n), and since we can permute the n − k “complementary indices”arbitrarily, we get
〈η, ζ〉g =1
k!
∑1≤
i1,...,ik≤n
η(Ei1 , . . . , Eik)ζ(Ei1 , . . . , Eik)
=1
k!(n−k)! · k!
∑σ∈S(n)
η(Eσ(1), . . . , Eσ(k))ζ(Eσ(1), . . . , Eσ(k))
=∑
σ∈S(k,n)
η(Eσ(1), . . . , Eσ(k))ζ(Eσ(1), . . . , Eσ(k)),
which is what we wanted to show.
Remark 2.10. The elements of S(k, n) are called (k, n)-shuffles.
Proposition 2.11. The basis
B = eα | α a k-dimensional multi-index with α1 < . . . < αk
of Λk(T ∗pM) is orthonormal with respect to 〈·, ·〉g.
Proof. Let eα, eβ ∈ B. Since eα(Eσ(1), . . . , Eσ(k)) = 1 if only if σ(i) = αi,1 ≤ i ≤ k, and otherwise equals zero, we have
〈eα, eβ〉g =∑
σ∈S(k,n)
eα(Eασ(1) , . . . , Eασ(k))eβ(Eασ(1) , . . . , Eασ(k))
= eα(Eα1 , . . . , Eαk)eβ(Eα1 , . . . , Eαk).
But eβ(Eσ(1), . . . , Eσ(k)) = 1 if only if σ(i) = βi, and otherwise equals zero,it must be the case that
〈eα, eβ〉g =
1 if α = β, that is, if eα = eβ0 otherwise
,
which is what we set out to prove.
Remark 2.12. Applying 〈, 〉g pointwise to differential forms yields the map
〈·, ·〉g : Ωk(M)× Ωk(M)→ C∞(M).
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Remark 2.13. In Section 3.2 we introduce an isomorphism ] : T ∗pM −→TpM that lets us define 〈·, ·〉g for covectors as
〈η, ζ〉g := 〈η], ζ]〉.
For decomposable elements η = η1∧ . . .∧ηk and ζ = ζ1∧ . . .∧ζk in Λk(T ∗pM),we can then define 〈·, ·〉g as
〈η1 ∧ . . . ∧ ηk, ζ1 ∧ . . . ∧ ζk〉g := det(〈ηi, ζj〉g). (2.2)
It is clear that the above definition coincides with our original definition forall basis vectors eα, eβ ∈ B, since we have
det(〈eαi , eβj 〉g) = det
δα1β1 0. . .
0 δαkβk
=
1 if α = β0 otherwise
.
A multilinearity argument then shows that (2.2) also coincides with theoriginal definition for arbitrary decomposable η and ζ, and from there itextends linearly to allow for indecomposable elements.
The importance of 〈·, ·〉g for our purposes lies in the fact that it lets usdefine the Hodge star operator on differential forms.
2.5 The Hodge star operator
Let (M, g) be a Riemannian manifold with boundary. We now introduce anautomorphism
? : Ω∗(M) −→ Ω∗(M),
known as the Hodge star operator, defined by requiring that for a k-form η,the identity
ζ ∧ η = 〈ζ, ?η〉Λk ωgholds for all ζ ∈ Ωn−k(M). It is linear over C∞(M) and has the propertythat its restriction to Ωk(M) is an isomorphism from the space of k-forms tothe space of (n− k)-forms, that is
?|Ωk(M) : Ωk(M) −→ Ωn−k(M).
Note that this makes sense since
dim Λk(T ∗M) =
(n
k
)=
(n
n− k
)= dim Λn−k(T ∗M).
We use the notation ?k when referring to the restriction of ? to Ωk(M), i.e.
?k := ?|Ωk(M).
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Proposition 2.14. Let (M, g) be a Riemannian manifold with boundary.The Hodge star operator is the unique automorphism on Ω∗(M) that maps thek-form η to the (n− k)-form ?η. Moreover, for each k ∈ 0, . . . , n, the map?k is an isomorphism from the space of k-forms to the space of n− k-formson M .
We use the following lemma, which is a finite-dimensional version of theRiesz representation theorem, in the proof of Proposition 2.14.
Lemma 2.15. Let V be a finite-dimensional vector space endowed with anondegenerate inner product g, and let f be a linear functional on V . Thenthere exists a unique vector v ∈ V such that
f(v) = g(v, w) for all w ∈ V .
Proof. To show uniqueness, assume that v exists. Let u1, . . . , un be anorthonormal basis for V , and write w = v1u1 + . . . vnun. We must havef(ui) = g(ui, v), and hence
v =n∑i=1
g(ui, ui)f(ui)ui =n∑i=1
f(ui)ui.
To show existence, we can easily check that v, defined as above, producesthe desired result.
Remark 2.16. In the proof of Lemma 2.15, we see that if we vary f smoothlyin V ∗, then v varies smoothly in V , and vice versa.
Proof of Proposition 2.14. Every top differential form on M can be writtenas f ωg for some smooth function f on M . Fix η ∈ Ωk(M). Then ζ ∧ η is atop form for all ζ ∈ Ωn−k(M), and thus
ζ ∧ η = fη(ζ)ωg, (2.3)
where fη(ζ) is smooth and
fη : Ωn−k(M) −→ C∞(M)
is linear over C∞(M). Moreover, fη is uniquely defined by (2.3), and therestriction of fη to a point is linear over R, i.e. at each point p ∈M the map
fη|Λn−k(T ∗pM) : Λn−k(T ∗pM) −→ R
is a linear functional uniquely determined by (2.3). This observation letsus use Lemma 2.15 to deduce that there exists a unique form θ ∈ Ωn−k(M)such that
fη(ζ) = 〈ζ, θ〉Λk for all ζ ∈ Ωk(M).
Take ?η := θ. Then ? is linear over C∞(M) and ker ? = 0, so it is indeed anautomorphism, and since ?η ∈ Ωn−k(M), its restriction to k-forms ?k is anisomorphism from Ωk(M) to Ωn−k(M). By Remark 2.16, θ is smooth.
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Our definition of ? is not very practical when it comes to actual computa-tions. Luckily for us, there exist equivalent definitions that lend themselvesmore easily to this task. The next proposition establishes several equivalentdefinitions of the Hodge star. We state it without proof.
Proposition 2.17. Let (M, g) be a Riemannian manifold with boundary.The following definitions of ? are equivalent.
1. Let η ∈ Ωk(M). Then ?η is defined by demanding that
ζ ∧ η = 〈ζ, ?η〉Λk ωg for all ζ ∈ Ωn−k(M).
2. Let η ∈ Ωk(M). Then ?η is defined by demanding that
ζ ∧ ?η = 〈ζ, η〉ωg for all ζ ∈ Ωk(M). (2.4)
3. Let e1, . . . , en be an orthonormal coframe defined on some open subsetU ∈M and let σ ∈ Sn. Then
?(eσ(1) ∧ . . . ∧ eσ(k)) := sgn(σ) eσ(k+1) ∧ . . . ∧ eσ(n). (2.5)
Since ? is linear, and since an orthonormal basis for Λk(T ∗M) on Ucan be found among the members of the set eσ(1)∧ . . .∧eσ(k) | σ ∈ Sn,this suffices to compute ?η for all η ∈ Ωk(M).
4. Let E1, . . . , En be an orthonormal frame defined on some open subsetU ∈ M and let η ∈ Ωk(M). Then ?η is defined on U to be the(n− k)-form for which
(?η)(Eσ(k+1), . . . , Eσ(n)) := sgn(σ) η(Eσ(1), . . . , Eσ(k)) (2.6)
for all σ ∈ Sn.
Remark 2.18. Definitions 3 and 4 in Proposition 2.17 work equally wellwith S(k, n) substituted for Sn.
Corollary 2.19. Let (M, g) be a Riemannian manifold with boundary. Then?1 = ωg.
Proof. This follows directly from definition 4 of Proposition 2.17,
Proposition 2.20. Applying ? twice yields
? ? η = (−1)k(n−k)η
for all η ∈ Ωk(M).
16
Proof. From definition 3 of Proposition 2.17 we know that for an orthonormalcoframe e1, . . . , en defined on some open subset U ∈ M and for σ ∈ Sn,we have
?(eσ(1) ∧ . . . ∧ eσ(k)) = sgn(σ) eσ(k+1) ∧ . . . ∧ eσ(n).
Applying the same definition of ? twice yields
? ? (eσ(1) ∧ . . . ∧ eσ(k)) = sgn(σ) ? (eσ(k+1) ∧ . . . ∧ eσ(n))
= sgn(σ)2(−1)x(eσ(1) ∧ . . . ∧ eσ(k))
= (−1)x(eσ(1) ∧ . . . ∧ eσ(k)).
where x = k(n− k) since we can get from
(σ(k + 1), . . . , σ(n), σ(1), . . . , σ(k))
to(σ(1), . . . , σ(n))
by k(n− k) adjacent transpositions (first move σ(1) n− k steps to the leftby adjacent transpositions, then σ(2) n − k steps to the left etc. until wehave moved σ(k); this adds up to k(n− k) adjacent transpositions).
Corollary 2.21. The inverse map of ? is given by
?−1 : Ω∗(M) −→ Ω∗(M)
η 7−→ (−1)k(n−k) ? η,(2.7)
i.e. ?−1 = ? in odd dimensions?−1 = (−1)k? in even dimensions.
Proof. The result follows easily by working backwards from (2.7) and usingProposition 2.20. First, we have
(−1)k(n−k) ? η = (−1)k(n−k) ?−1 ? ? η = (−1)2k(n−k) ?−1 η = ?−1η.
Then it is straightforward to verify that
k(n− k) = nk − k2 ≡
0 (mod 2) if n is oddk (mod 2) if n is even
.
Proposition 2.22. Consider R3 with the standard metric, with x, y and zthe standard coordinates. Then the Hodge star operator on Ω∗(R3) is givenby
?1 = dx ∧ dy ∧ dz ?(dx ∧ dy ∧ dz) = 1
?dx = dy ∧ dz ?(dy ∧ dz) = dx
?dy = −dx ∧ dz ?(dx ∧ dz) = −dy
?dz = dx ∧ dy ?(dx ∧ dy) = dz.
17
Proof. Note that since ? is linear, it is sufficient to determine where it takesthe (standard) basis vectors of Ωk(R3) for 0 ≤ k ≤ 3. The basis vectors are
1 for Ω0(R3),
dx, dy and dz for Ω0(R3),
dx ∧ dy, dx ∧ dz and dy ∧ dz for Ω0(R3),
dx ∧ dy ∧ dz for Ω0(R3).
As we have already seen, these bases are orthonormal with respect to tothe inner product on k-forms 〈·, ·〉g. The result thus follows immediately
from Proposition 2.17. For example, let σ =
(1 2 32 1 3
)∈ S3. Using cycle
notation we can write σ = (1 2)(3), hence sgn(σ) = (1 2)(3). Let (e1, e2, e3)denote the standard coframe, i.e. e1 = dx, e2 = dy and e3 = dz. Then
?dy = ?e2 = ?eσ(1)
= sgn(σ) eσ(2) ∧ eσ(3)
= −e1 ∧ e3 = −dx ∧ dz
and
?(dx ∧ dy) = ?(−dy ∧ dx)
= − ? (dy ∧ dx)
= − ? (e2 ∧ e1)
= − ? (eσ(1), eσ(2))
= −sgn(σ) eσ(3) = e3 = dz.
The rest of the proof follows from similar calculations.
2.6 An L2-inner product on differential forms
Let M be a compact smooth orientable manifold. We define an L2-innerproduct on the space differential k-forms on M as
〈·, ·〉 : Ωk(M)× Ωk(M) −→ R
(η, ζ) 7−→∫Mη ∧ ?ζ.
That this defines an inner product on Ωk(M) is easy to verify: since we canwrite η ∧ ?ζ = 〈η, ζ〉g, ωg, the required properties of symmetry, linearity and
18
positive-definiteness all follow from the corresponding properties of 〈·, ·〉g.along with the linearity of the integral.
Equipped with this inner product, the vector space Ωk(M) becomes aninfinite-dimensional inner product space.
Proposition 2.23. Let (M, g) be a compact Riemannian manifold withboundary. The Hodge star operator preserves the L2-inner product in thesense that
〈?η, ?ζ〉 = 〈η, ζ〉for all η, ζ ∈ Ω(M)p.
Proof. The result is obtained by a straighforward computation:
〈?η, ?ζ〉 =
∫M?η ∧ ? ? ζ = (−1)2p(n−p)
∫Mζ ∧ ?η = 〈ζ, η〉 = 〈η, ζ〉.
2.7 The codifferential
Let (M, g) be a Riemannian manifold with boundary. The codifferential,denoted d∗, is the map defined by
d∗ : Ωk(M) −→ Ωk−1(M)
η 7−→ (−1)n(k−1)+1 ? d ? η(2.8)
for each k. Since ? is used twice in the definition, the codifferential does notdepend on the orientation of M .
Proposition 2.24. Let (M, g) be a Riemannian manifold with boundary.An alternative expression for the codifferential is given by
d∗η = (−1)k ?−1 d ? η
for all η ∈ Ωk(M).
Proof. Let η ∈ Ωk(M). Then d ? η ∈ Ωn−k+1(M). Using Corollary 2.21 wesee that
?(d ? η) = (−1)(n−k+1)(n−(n−k+1)) ?−1 (d ? η)
= (−1)kn−k−n−1 ?−1 (d ? η).
Substituting the above expression into (2.8) yields
d∗η = (−1)n(k−1)+1 ? d ? η
= (−1)n(k−1)+1(−1)kn−k−n−1 ?−1 d ? η
= (−1)k ?−1 d ? η,
which is what we wanted to prove.
19
Proposition 2.25. Let (M, g) be a Riemannian manifold with boundary.The codifferential is nilpotent of degree 2, that is
d∗d∗η = 0
for all η ∈ Ω∗(M).
Proof. Let η ∈ Ωk(M). Using Proposition 2.24 along with the fact thatdd = 0, we get
d∗d∗η = (−1)k−1 ?−1 d ? (−1)k ?−1 d ? η = − ?−1 dd ? η = 0.
The following proposition gives an important property of the codifferentialthat is true only for Riemannian manifolds without boundary.
Proposition 2.26. Let (M, g) be a compact Riemannian manifold withoutboundary. Then
〈dη, ζ〉 = 〈η, d∗ζ〉
for all η ∈ Ωk(M) and ζ ∈ Ωk+1(M).
Proof. Let η ∈ Ωk(M) and ζ ∈ Ωk+1(M). Using the identity
dη ∧ ?ζ = d(η ∧ ?ζ)− (−1)k(η ∧ d ? ζ)
(recall that this is one of the defining properties of the exterior derivative)and the alternative characterization of d∗ in Proposition 2.24, we compute
〈dη, ζ〉 =
∫Mdη ∧ ?ζ
=
∫M
(d(η ∧ ?ζ)− (−1)k(η ∧ d ? ζ)
)=
∫Md(η ∧ ?ζ) +
∫Mη ∧ (?(−1)k+1 ?−1 d ? ζ)
=
∫∂M
η ∧ ?ζ +
∫Mη ∧ ?d∗ζ = 〈η, d∗ζ〉,
where the last equality holds because ∂M is empty.
Remark 2.27. In light of Proposition 2.26, the notation d∗ makes perfectsense whenever M is compact without boundary, since then d∗ is the L2-adjoint of d. In this case, the condition that 〈dη, ζ〉 = 〈η, d∗ζ〉, can be takenas the definition of d∗, and the fact that d∗ can be expressed in a neat wayusing the Hodge star will then come as a nice surprise.
20
When M has nonempty boundary it is no longer the case that d and d∗
are adjoint operators, since the leftmost integral in∫∂M
η ∧ ?ζ =
∫Md(η ∧ ?ζ) = 〈dη, ζ〉 − 〈η, d∗ζ〉
is in general non-zero, but is zero when ı∗ζ = 0 or ı∗ ? η = 0, where ı∗ is thepullback by the inclusion map ı : ∂M −→ M . For this reason it might besensible to use a different symbol than d∗ when referring to the codifferentialon a manifold with nonempty boundary. It is common to let δ denote thecodifferential instead of d∗, but this alternative notation does not alwaysseem to be motivated by the above considerations.
In any case, throughout this text we will use d∗ to denote the codifferential,even when it is not the L2-adjoint of d.
21
3 Vector calculus and differential forms
In this section we investigate how differential forms and vector calculusare related. In particular, we see how the exterior derivative d and thecodifferential d∗ extends the classical gradient, curl and divergence of vectorcalculus. As an application, we derive the well known Green’s theorem,divergence theorem and Kelvin-Stokes theorem encountered in vector calculus,from Stokes’ theorem (Theorem 2.7).
3.1 The Laplace-de Rham operator
Let (M, g) be a Riemannian manifold with boundary. The Laplace-de Rhamoperator ∆ on differential k-forms on M is defined by
∆ : Ωk(M) −→ Ωk(M)
η 7−→ (d+ d∗)2(η).
Hence, for a k-form η we have
∆η = (d+ d∗)2(η) = (dd∗ + dd∗d)(η) = dd∗η + d∗dη
Proposition 3.1. The Laplace-de Rham operator ∆ commutes with ?, dand d∗
Proof. Let (M, g) be a Riemannian manifold with boundary and let η ∈Ωk(M).
We first show that ∆ commutes with ?. By expanding the definitionsand rearranging, we get
?(dd∗η) = ?d(−1)n(k−1)+1 ? d ? η
= (−1)n(k−1)+1 ? d ? (d ? η)
= (−1)n((n−k+1)−1)+1 ? d ? (d ? η)
= d∗d(?η)
22
and
?(d∗dη) = ?(−1)n((k+1)−1)+1 ? d ? dη
= ? ? d(−1)nk+1 ? d(−1)k(n−k) ? (?η)
= (−1)(n−k)(n−(n−k))d(−1)k+1 ? d ? (?η)
= d(−1)nk+1 ? d ? (?η)
= d(−1)n((n−k)−1)+1 ? d ? (?η)
= dd∗(?η),
so that
?∆η = ?(dd∗η + d∗dη) = ?(dd∗η) + ?(d∗dη) = ∆ ? η,
i.e. ∆ commutes with ?.To show that d∆η = ∆dη, we compute
d∆η = d(dd∗η + d∗dη)
= ddd∗η + dd∗dη
= dd∗dη + d∗d∗dη = ∆dη.
The proof that d∗∆η = ∆d∗η is analogous.
When M is compact, the Laplace-de Rham operator combines the twoconditions of closedness (dη = 0) and coclosedness (d∗η = 0) into a singlecondition. The following proposition spells out the details.
Proposition 3.2. Let η be a differential form on a closed Riemannianmanifold. Then ∆η = 0 if and only if dη = 0 and d∗η = 0.
Proof. Let M denote our manifold. Since M is compact, every differentialform has compact support. Hence the inner product of two k-forms isalways defined. Moreover, M has no boundary, which implies that d∗ is theL2-adjoint of d. Using these facts we compute
〈∆η, η〉 = 〈dd∗η + d∗dη, η〉 = 〈dd∗η, η〉+ 〈d∗dη, η〉
= 〈d∗η, d∗η〉+ 〈dη, dη〉 = ||d∗η||2 + ||d∗η||2,
which shows that ∆η = 0 implies that dη = 0 and d∗η = 0.The other direction is trivial.
23
3.2 Musical isomorphisms
Let M be a smooth oriented manifold equipped with a Riemannian metric g.The isomorphism
[ : TpM −→ T ∗pM
defined by requiring that
v[(w) = gp(v, w) ∀v, w ∈ TpM,
and its inverse] : T ∗pM −→ TpM,
defined bygp(ω
], v) = ω(v) ∀ω ∈ T ∗pM, v ∈ TpM,
take smooth vector fields to smooth covector fields and vice versa whenapplied pointwise to TM and T ∗M . Hence, by applying [ to smooth vectorfields and ] to smooth covector fields, we get the inverse isomorphisms
[ : Γ(TM) −→ Ω1(M)
] : Ω1(M) −→ Γ(TM)
defined pointwise by
V [(W ) = gp(V,W ) ∀V,W ∈ Γ(TM)
gp(ω], V ) = ω(V ) ∀ω ∈ Ω1(M), V ∈ Γ(TM).
(See Lemma 2.15 and Remark 2.16.)Next, we want to establish an isomorphism β from the tangent bundle to
the bundle of alternating covariant (n− 1)-tensors on M . This is achievedby composing ? and [ as follows.
β : TM −→ Λn−1T ∗M
v 7−→ ?v[
The inverse of β is given by
β−1 = (−1)n+1] ? .
Applying β to smooth vector fields and β−1 to smooth n− 1-forms, we getthe inverse isomorphisms
β : Γ(TM) −→ Ωn−1(M)
β−1 : Ωn−1(M) −→ Γ(TM)
24
defined by
β(V ) = ?V [ and β−1(η) = (−1)n+1(?η)].
The interior product on differential forms provides an alternative wayof expressing β, which follows from the next proposition and is stated asCorollary 3.4.
Proposition 3.3. Let (M, g) be an oriented Riemannian n-manifold, letV ∈ Γ(TM) and let η ∈ Ωk(M). Then
?iV (η) = (−1)k−1(V [ ∧ ?η). (3.1)
Proof. Let p ∈ M and let E1, . . . , En be a local orthonormal at p, withe1, . . . , en the corresponding coframe, so that
V (p) =
n∑i=1
viEi and η(p) = f(p) eπ(1) ∧ . . . ∧ eπ(k),
where π ∈ Sn and because both sides are linear, without loss of generality, ηhas been chosen to be decomposable into a wedge product of basis 1-forms.As all the operations involved in the identity (3.1) are defined pointwise, wewill work exclusively in TpM and T ∗pM throughout the rest of this proof,and so we will omit writing p in calculations in order to improve legibility,e.g. we will write f instead of f(p) and V instead of V (p) as the referenceto p is understood.
Expanding the left hand side of (3.1) yields
iV (η) = fk∑i=1
(−1)i−1 eπ(i)(V ) eπ(1) ∧ . . . ∧ eπ(i) ∧ . . . ∧ eπ(k)
= fk∑i=1
(−1)i−1 vπ(i) eπ(1) ∧ . . . ∧ eπ(i) ∧ . . . ∧ eπ(k).
Next, from the definition of ? we see that
?η = sgn(π)f eπ(k+1) ∧ . . . ∧ eπ(n).
Since we are working in an orthonormal frame, we have V [ =n∑i=1
viei. A
straightforward computation then shows that
V [ ∧ ?η = sgn(π)f
n∑i=1
viei ∧ eπ(k+1) ∧ . . . ∧ eπ(n)
= sgn(π)fk∑i=1
vπ(i)eπ(i) ∧ eπ(k+1) ∧ . . . ∧ eπ(n).
25
Applying ? to iV (η) introduces a factor sgn(τik π) = (−1)i−ksgn(π), so that
?iV (η) = (−1)k−1 sgn(π)fk∑i=1
vπ(i)eπ(i) ∧ eπ(k+1) ∧ . . . ∧ eπ(n)
= (−1)k−1(V [ ∧ ?η),
which is the sought identity.
Corollary 3.4. Let (M, g) be an oriented Riemannian n-manifold and letV ∈ Γ(TM). Then
β(V ) = iV (ωg).
Proof. Take η = ωg in Proposition 3.3. Then k = n, ?η = 1 and V [∧?η = V [.Applying ? to (3.1) and noting that ?? = (−1)k(n−k), we get the left handside
? ? iV (ωg) = (−1)(k−1)(n−(k−1))iV (ωg) = (−1)n−1iV (ωg),
which, when put together with the right hand side, yields
(−1)n−1iV (ωg) = (−1)n−1 ? V [ ⇐⇒ iV (ωg) = ?V [.
Since β(V ) = ?V [, we are done with the proof.
3.3 Gradient and divergence
Let (M, g) be a Riemannian manifold. We define a map grad by
grad : C∞(M) −→ Γ(TM)
f 7−→ (df)].(3.2)
For f ∈ C∞(M), we call the vector field grad (f) the gradient vector field (orjust the gradient) of f . By definition, for V ∈ Γ(TM) we have 〈 grad f, V 〉 =df(V ) = V f . Thus 〈 grad f, V 〉 gives the directional derivative of f along V ,and grad f is the unique vector field with this property.
Moving on, we define another map, div , by
div : Γ(TM) −→ C∞(M)
V 7−→ ?−1(dβ(V )).(3.3)
The div map does not depend on the orientation of the manifold. Indeed,if e1, . . . en is an orthonormal coframe on M , then any odd permutationπ ∈ Sn yields a differently oriented orthonormal coframe eπ(1), . . . , eπ(n) forwhich ?(eπ(1)∧ . . .∧eπ(n)) = −? (e1∧ . . .∧en). This follows from Proposition2.17. From Corollary 3.4 we see that β changes sign in the same way underthis change of coframe. In the definition of div , these two sign changes
26
cancel each other, implying that div is defined invariantly on all Riemannianmanifolds, regardless of orientation.
The codifferential, eager not to be left behind, can also be used to definegrad and div . This will prove to be very useful later on, and the followingproposition shows how it can be done.
Proposition 3.5. Let (M, g) be a Riemannian manifold and let f ∈ C∞(M)and V ∈ Γ(TM). Then the following statements are true:
1. grad f = −β−1(d∗ ? f).
2. divV = −d∗V [.
Proof.
1. Since ?f is an n-form and d∗ ? f is an (n− 1)-form we have
−β−1(d∗ ? f) = (−1)n[?(−1)n ?−1 d ? (?f)]]
= [df ]] = grad f.
2. Since V [ is a 1-form we have
−d∗V [ = (−1)1+1 ?−1 d ? V [
= ?−1dβ(V ) = divV.
3.4 Euclidean space
We will use the term Euclidean n-space, or just n-space, to denote Rnequipped with the Euclidean metric. If the dimension is unimportant orclear from context, we might drop the n and write Euclidean space instead.
Proposition 3.6. The following statements are true in Euclidean n-spacewith standard coordinates x1, . . . , xn:
1. Let V =n∑i=1
Vi∂
∂xi∈ Γ(TRn) and η =
n∑i=1
ηidxi ∈ Ω1(Rn). Then
V [ =
n∑i=1
Vidxi and η] =
n∑i=1
ηi∂
∂xi. (3.4)
2. Let f ∈ C∞(Rn). Then
grad f =
n∑i=1
∂f
∂xi
∂
∂xi.
27
3. Let V =
n∑i=1
Vi∂
∂xi∈ Γ(TRn). Then
divV =
n∑i=1
∂Vi∂xi
.
4. Let V,W ∈ Γ(TRn). Then V ·W = ?(V [ ∧ ?W [), where · denotes theusual dot product.
Proof.
1. In standard coordinates on Rn we have g = δijdxidxj . Hence
V [(W ) = gp(V,W ) = V ·W,
which implies that V [ =n∑i=1
Vidxi. From this we also conclude that
η] =
n∑i=1
ηi∂
∂xi.
2. It suffices to observe that
df =n∑i=1
∂f
∂xidxi.
The result then follows from (3.4).
3. From the definitions we see that (divV )ωg = d ? V [. From there wecompute
d(?V [) = d ?n∑i=1
(Vi dxi) = dn∑i=1
?(Vi dxi)
= dn∑i=1
((−1)i−1Vi dx1 ∧ . . . ∧ dxi ∧ . . . ∧ xn
)
=n∑i=1
(dVidxi
dx1 ∧ . . . ∧ xn)
=
(n∑i=1
dVidxi
)dx1 ∧ . . . ∧ xn,
from which the desired conclusion follows.
28
4. We compute
V [ ∧ ?W [ =n∑i=1
Vi dxi ∧ ?n∑i=1
Wi dxi
=
n∑i=1
Vi dxi ∧n∑i=1
(−1)i−1Wi dx1 ∧ . . . ∧ dxi ∧ . . . ∧ dxn
=
n∑i=1
(−1)i−1ViWi dxi ∧ (dx1 ∧ . . . ∧ dxi ∧ . . . ∧ dxn)
=
(n∑i=1
ViWi
)ωg,
and thus ?(V [ ∧ ?W [) = V ·W .
From the above proposition we see that our definitions of grad and divgeneralize the classical gradient and divergence functions. We will denotethese latter two functions by ∇ and ∇· respectively, in order to distinguishthem from grad and div . Thus, in Euclidean space we have
∇f =n∑i=1
∂f
∂xi
∂
∂xiand ∇ · V =
n∑i=1
∂Vi∂xi
,
so that the diagrams
C∞(Rn) Γ(TRn)
Ω0(Rn) Ω1(Rn)
∇
id [
d
and
Γ(TRn) C∞(Rn)
Ωn−1(Rn) Ωn(Rn)
∇·
β ?
d
(3.5)
commute. In the special case that n = 2 we have Ω1(R2) = Ω2−1(R2), sothat [ and β have the same domain and codomain, which suggests that wemay combine the two diagrams into one. But [ 6= β in the two dimensionalcase, so in order to combine the diagrams we need to choose either [ or β asour preferred map and then adjust either ∇ or ∇· accordingly. Let’s say wechoose [. We then get the commutative diagram
C∞(R2) Γ(TR2) C∞(R2)
Ω0(R2) Ω1(R2) Ω2(R2),
∇×
id
F
[ ?
d d
29
where F = (∇·) rot, and rot is the linear map that rotates vector fields by
sending∂
∂xto
∂
∂yand
∂
∂yto − ∂
∂x. In 3-space, on the other hand, we need
the classical curl operator on vector fields to glue the two diagrams together.Similarly, in light of Proposition 3.5 we see that the diagrams
C∞(Rn) Γ(TRn)
Ωn(Rn) Ωn−1(Rn)
∇
? β
−d∗and
Γ(TRn) C∞(Rn)
Ω1(Rn) Ω0(Rn)
∇·
[ id
−d∗(3.6)
commute. Again, when n = 2 it is possible adjust them so that they combine,and when n = 3 the curl operator lets us join them together.
3.5 Three-space
We now turn our attention to Euclidean 3-space, where the standard coordi-nates will be denoted x, y and z.
In 3-space, we define the curl operator on vector fields by
curl : Γ(TM) −→ Γ(TM)
V 7−→ β−1dV [.
The definition of curl only works for 3-dimensional manifolds, since dV [ isa 2-form and β−1 takes (n− 1)-forms to vector fields, so that β−1dV [ onlymakes sense when n = 3.
Remark 3.7. In fact, our definition of curl works well on any orientedRiemannian 3-manifold. We will use curl in this more general setting inTheorem 3.14.
Let ∇× and × denote the classical curl operator and cross product onR3. The following proposition shows that curl coincides with ∇× in R3.
Proposition 3.8. Let V and W be smooth vector fields on R3 and letf ∈ C∞(R3). Then the following statements are true:
1. ∇× V = curlV = (d∗βV )].
2. V ×W = [?(V [ ∧W [)]].
3. curl grad f = 0 and div curlV = 0.
Proof.
30
1. We begin by showing that ∇× V = β−1(dV [). First we compute
dV [ =∂V1
∂ydy ∧ dx+
∂V1
∂zdz ∧ dx+
∂V2
∂xdx ∧ dy +
∂V2
∂zdz ∧ dy
+∂V3
∂xdx ∧ dz +
∂V3
∂ydy ∧ dz
=
(∂V3
∂y− ∂V2
∂z
)dy ∧ dz +
(∂V1
∂z− ∂V3
∂x
)dx ∧ dz
+
(∂V2
∂x− ∂V1
∂y
)dx ∧ dy.
Since n = 3 we have β−1 = (−1)n+1]? = ]?, where ?(dy ∧ dz) = dx,?(dx ∧ dz) = −dy and ?(dx ∧ dy) = dz. From this we conclude that∇× V = β−1dV [.
Next we show that ∇ × V = (d∗βV )]. Using the same identities asabove, we have
d∗βV = (−1)3(2+1)+1 ? d ? (V1 dy ∧ dz − V2 dx ∧ dz + V3 dx ∧ dy)
= ?d(V1 dx+ V2 dy + V3dz)
= ?
[(∂V3
∂y− ∂V2
∂z
)dy ∧ dz +
(∂V1
∂z− ∂V3
∂x
)dx ∧ dz
+
(∂V2
∂x− ∂V1
∂y
)dx ∧ dy
].
After applying ] to the above expression we end up in the exact samesituation as above. Again, this is because β−1 = ]?. And so we aredone.
2. We first compute
V [ ∧W [ = (V1dx+ V2dy + V3dz) ∧ (W1dx+W2dy +W3dz)
= V1W2 dx ∧ dy + V1W3 dx ∧ dz + V2W1 dy ∧ dx
+ V2W3 dy ∧ dz + V3W1dz ∧ dx+ V3W2dz ∧ dy
= (V2W3 − V3W2) dy ∧ dz + (V1W3 − V3W1) dx ∧ dz
+ (V1W2 − V2W1) dx ∧ dy.
31
Applying ? to the above expression yields
[?(V [ ∧W [)]] = [(V2W3 − V3W2) dx− (V1W3 − V3W1) dy
+ (V1W2 − V2W1) dz]]
= (V2W3 − V3W2)∂
∂x− (V1W3 − V3W1)
∂
∂y
+ (V1W2 − V2W1)∂
∂z
= V ×W,
which is what we set out to prove.
3. Unwinding the definitions and using that d2 = 0 we see that
curl grad f = β−1d((df)])[ = β−1ddf = 0
anddiv curlV = ?−1dβ(β−1dV [) = ?−1ddV [ = 0.
Proposition 3.8 tells us that the diagrams
Γ(TR3) Γ(TR3)
Ω1(R3) Ω2(R3)
∇×
[ β
d
and
Γ(TR3) Γ(TR3)
Ω2(R3) Ω1(R3)
∇×
β [
d∗
(3.7)
commute. Thus, in 3-space we can combine the diagrams (3.5) with (3.7),which together with the fact that in 3-space we have
?−1 = ? and d∗ = (−1)k ? d? (3.8)
lets us draw the following commutative diagram with exact rows and iso-morphisms in the columns, which nicely sums up much of our work sofar.
C∞(R3) Γ(TR3) Γ(TR3) C∞(R3)
Ω0(R3) Ω1(R3) Ω2(R3) Ω3(R3)
Ω3(R3) Ω2(R3) Ω1(R3) Ω0(R3)
C∞(R3) Γ(TR3) Γ(TR3) C∞(R3)
∇
id
∇×
[
∇·
β ?
d
?
d
?
d
? ?
−d∗
?
d∗
β−1
−d∗
] id
∇ ∇× ∇·
(3.9)
32
3.6 Vector calculus identities
Diagram (3.9) shows us how the d operator generalizes vector calculus oncewe figure out the appropriate ways to translate between vector fields anddifferential forms. In this section, we use the results of the previous sectionsto showcase how the theory of differential forms lets us deduce and reinterpretsome familiar vector calculus identities. Recall that we already proved that∇×∇ = 0 and div∇× = 0 in Proposition 3.8. Throughout the rest of thissection we work in Euclidean space. For a self-contained treatment of thismaterial, including a discussion about applications to electodynamics, seeSchleifer’s article [11].
We start off with a few first order identities, i.e. identities involving onlyfirst derivatives.
Proposition 3.9 (First order identities). Let f ∈ C∞(R3) and V,W ∈Γ(TR3). Then the following identities hold:
1. ∇(fg) = f∇g + g∇f
2. ∇× (fV ) = f(∇× V ) + (∇f)× V
3. ∇ · (fV ) = f(∇ · V ) + V · (∇f)
Proof.
1. Since d is the differential for 0-forms, we have
∇(fg) = (d(fg))] = (fdg + gdf)].
2. Using that d(fV )] = d(fV ]) = df ∧ V ] + fdV ], we compute
∇× (fV ) = β−1(d(fV )]) = [?(d(fV )])]]
= [?(df ∧ V ])]] + [f ? (dV ]))]]
= (∇f)× V + f(∇× V )).
3. In the following computation we use that ? = ?−1 in 3-space.
∇ · (fV ) = ?−1d(β(fV )) = ?−1d(fβ(V ))
= ?−1(df ∧ ?V [ + fdβ(V ))
= ?(df ∧ ?V [) + ?−1(fdβ(V ))
= V · (∇f) + f(∇ · V ).
33
Second order identities, i.e. identities involving second derivatives, arekeen not to be left out, so we include one such identity here. Let ∇2 denotethe classical Laplacian, defined by
∇2 : C2(Rn) −→ C0Rn)
f 7−→ ∇ · ∇f.
The following proposition shows how the Laplace-de Rham operator general-izes the classical Laplacian.
Proposition 3.10. Let f ∈ C∞(Rn). Then ∆f = −∇2f .
Proof. We compute
∆f = dd∗f + d∗df = d(−1)3(0−1)+1 ? d ? f + d∗df
= d∗ ? df
= d∗((df)])[
= −∇ · (∇f) = −∇2f.
3.7 Classical theorems
Stokes’ theorem (Theorem 2.7) generalizes the well known Green’s theorem,divergence theorem and Kelvin-Stokes theorem. We state and prove themin this section. Whereas Green’s theorem is a more or less immediateconsequence of Stoke’s theorem, we rely on a lemma to prove the other two.
Theorem 3.11 (Green’s theorem). Let D ⊂ R2 be a compact regular domain,and let η = P dx+Qdy be a 1-form on D. Then∫
D
(∂P
∂y− ∂Q
∂x
)dx dy =
∫∂D
P dx+Qdy. (3.10)
Proof. We have
dη =∂P
∂ydx ∧ dy +
∂Q
∂xdy ∧ dx =
(∂P
∂y− ∂Q
∂x
)dx ∧ dy.
Applying Stokes’ theorem yields∫D
(∂P
∂y− ∂Q
∂x
)dx ∧ dy =
∫Ddη =
∫∂D
η.
From here the theorem follows from the definition of integration of differentialforms over domains of integration in Rn.
34
Lemma 3.12. Let (M, g) be an oriented Riemannian n-manifold with bound-ary, let g be the induced metric on ∂M and let N be the outward-pointingunit normal vector field along ∂M . Then, for any vector field V along ∂M ,we have
ı∗(β(V )) = g(V,N)ωg,
where ı : ∂M −→M is the inclusion map of the boundary.
Proof. Let V = VT + VN , where VN = g(V,N)N and VT = V − VN . Thatis, VT and VN are the components of V that are respectively tangent andnormal to ∂M . Thus we get
ı∗(β(V )) = ı∗(βVT + βVN ) = ı∗(β(VT )) + ı∗(β(VN )).
Recall that by Corollary 3.4 we have β(V ) = iV (ωg). Hence, for vectorsV1, . . . , Vn−1 tangent to ∂M we have
β(VT )(V1, . . . , Vn−1) = (iVTωg)(V1, . . . , Vn−1)
= ωg(VT , V1, . . . , Vn−1) = 0,
where the last equality holds because the tangent space of ∂M is (n − 1)-dimensional, and thus a set of n vectors that are tangent to ∂M must belinearly dependent. It follows that
ı∗(β(VT )) = 0.
Consider now ı∗(β(VN )). Writing VN = g(V,N)N and using the linearityof the pullback operator, we get
ı∗(β(g(V,N)N)) = g(V,N)ı∗(β(N)).
Now ı∗(β(N)) = ı∗(iNωg) is an orientation form on ∂M corresponding tothe orientation induced by N . Let p ∈ ∂M and let (E1, . . . , En−1) be anorthonormal frame of ∂M at p. Then we get a positively oriented orthonormalframe (N(p), E1(p), . . . , En−1(p)) for M at p, and thus
ı∗(β(N))(E1, . . . , En−1) = ı∗(iNωg)(E1, . . . , En−1)
= ωg(N,E1, . . . , En−1) = 1.
It follows that ı∗(β(N)) = ωg, and hence
ı∗(β(VN )) = g(V,N)ı∗(β(N)) = g(V,N)ωg,
which is what we wanted to prove.
35
Theorem 3.13 (The divergence theorem). Let (M, g) be an oriented Rie-mannian n-manifold with boundary, let g be the induced metric on ∂M ,let N be the outward-pointing unit normal vector field along ∂M and letV ∈ Γ(TM) be compactly supported. Then∫
M( divV )ωg =
∫∂M
g(V,N)ωg.
Proof. By definition we have ? divV = dβ(V ), and since dβ(V ) is an n-form,we have
dβ(V )ωg = ?dβ(V ) = ? ? divV = divV.
Applying Stokes’ theorem yields∫M
( divV ) ωg =
∫Mdβ(V ) =
∫∂M
ı∗β(V ), (3.11)
where ı : ∂M −→M is the inclusion map, and from Lemma 3.12 we have
ı∗(β(V )) = g(V,N)ωg. (3.12)
The divergence theorem is now obtained by combining (3.11) and (3.12).
Theorem 3.14 (Kelvin-Stokes theorem). Let (Y, g) be an oriented Rieman-nian 3-manifold with boundary and let S ⊂ Y be a compact oriented smooth2-dimensional submanifold with boundary. Let N be the unit normal vectorfield along S that determines its orientation and let T be the unique positivelyoriented unit tangent vector field along ∂S. Let V ∈ Γ(TY ). Then∫
Sg( curlV,N)ωg =
∫∂Sg(V, T )ωg,
where g and g are the metrics induced by g on S and ∂S respectively.
Proof. By slightly rearranging our definition of curl , we get
β( curlV ) = dV [,
and by Stokes’ theorem we then have∫Sd V [ =
∫Sı∗(d V [) =
∫∂S∗V [ =
∫∂SV [, (3.13)
where ı : S −→ M and : ∂S −→ S are the inclusion maps. From Lemma3.12 we see that
ı∗dV [ = ı∗(β( curlV )) = g( curlV,N)ωg. (3.14)
36
We now turn our attention to ∗ V [. It is a top-degree form on ∂S, so itmust be of the form f ωg for some f ∈ C∞(∂S). As T is the unique positivelyoriented unit tangent vector field on ∂S, we know that ωg(T ) = 1. Hence
f = f ωg(T ) = (∗ V [)(T ) = V [(T ) = g(V, T ),
and therefore∗V [ = g(V, T )ωg. (3.15)
Lastly, substituting (3.14) and (3.15) into (3.13) completes the proof.
37
4 Algebraic topology
In this section, we briefely introduce some notions and results from algebraictopology that will prove useful later on. In particular, we introduce singularhomology and cohomology, as well as reduced and relative homology, and deRham cohomology. I refer the reader to Hatcher, [6], and Bott and Tu, [2],for a tourough treatment of this material.
4.1 Singular homology and cohomology
The standard n-simplex, denoted ∆n, is the convex combination of thestandard basis vectors e0, . . . , en in Rn+1, i.e.
∆n :=
(x0, . . . , xn) ∈ Rn+1
∣∣∣∣ x0, . . . , xn ≥ 0 and
n∑i=0
xi = 1
.
Let X be a topological space. A continuous mapping σk : ∆k −→ X is calleda singular k-simplex. Let R be a unital ring. (From this point onwards, weassume all rings to be unital.) We let Ck(X;R) denote the group of formalsums of k-simplexes with coefficients in R, and call the elements of Ck(X;R)singular n-chains. Define a (linear) homomorphism
∂k : Ck(X;R) −→ Ck−1(X;R),
called the boundary operator, as follows: whenever σk = (σk0 , . . . , σkk) is a
k-simplex, i.e. a basis vector of Ck(X), let
∂kσk :=
∑i
(−1)i(σk0 , . . . , σki , . . . , σ
kk).
The boundary operator has the property that ∂k−1∂k = 0, and hence thediagram
. . . −→ C2(X;R)∂2−→ C1(X;R)
∂1−→ C0(X;R)∂0−→ 0, (4.1)
is a chain complex, called the singular chain complex. The k-th singularhomology group of X with coefficients in R, denoted Hk(X;R), is defined tobe the k-th homology group of this complex, i.e.
Hk(X;R) :=ker ∂k
im ∂k+1.
38
Remark 4.1. Since we take our coefficients from a ring R, the homologygroups that we just defined are actually R-modules. If we take the coefficientsfrom an Abelian group instead of a ring, the homology groups become groups.
Consider now the singular chain complex
. . . −→ C2(X;Z)∂2−→ C1(X;Z)
∂1−→ C0(X;Z)∂0−→ 0. (4.2)
Here, the coefficients are integers, and thus the group Ck(X;Z) can bethought of as being the free abelian group on the set of singular k-simplices.Take R to be a ring, and dualize (4.2) by applying the functor Hom(·, R) toobtain the cochain complex
0 −→ C0(X;R)∂0−→ C1(X;R)
∂1−→ C2(X;R) −→ . . . ,
where Ck(X;R) := Hom(Ck, R) and ∂k := Hom(∂k, R). The k-th singularcohomology group of X with coefficients in R, denoted Hk(X;R), is definedto be the quotient
Hk(X;R) :=ker ∂k
im ∂k−1.
Remark 4.2. We will oftentimes be lazy and refer to singular homology asjust homology and singular cohomology as just cohomology. When R = Z,one oftentimes simply writes Hk(X) instead of Hk(X;R) and Hk(X) insteadof Hk(X;R).
Remark 4.3. The elements of ker ∂0 = C0(X;R) are formal sums of con-tinuous maps from point x0 = 1 ∈ R1 to X, whereas im ∂1 is generatedby elements of the form a − b, where a, b ∈ ∆0 → Xi and Xi denotesa path component of X. From this it can be seen that the quotientH0(X;R) = ker ∂0/ im ∂1 is isomorphic the free R-module over the pathcomponents of X. If the number of path components is n < ∞, thenH0(X;R) ' Rn.
We also need to consider reduced homology and relative homology. Letε : C0(X;R) −→ R be the map given by
σ
(∑i
niσi
):=∑i
ni
and define the k-th reduced homology of X with coefficient in R, denoted byHk(X), to be the k-th homology group of the chain complex
. . . −→ C2(X;R)∂2−→ C1(X;R)
∂1−→ C0(X;R)ε−→ R −→ 0.
39
The k-th homology group of X relative to a subspace Y ⊂ X, denoted byHk(X,Y ;R) is defined to be the k-th homology group of the quotient chaincomplex
. . . −→ C2(X;R)
C2(Y ;R)
∂2−→ C1(X;R)
C1(Y ;R)
∂1−→ C0(X;R)
C0(Y ;R)
∂0−→ 0,
where the boundary map ∂i is the usual boundary map that has descendedto a map on the quotient Ci(X;R)/Ci(Y ;R).
4.2 Poincare-Lefschetz duality
In this section we prove that for any compact topological manifold M withboundary and for any field F, we have Hk(M ;F) ' Hn−k(M,∂M ;F) andHk(M,∂M ;F) ' Hn−k(M ;F), This follows from the Poincare-Lefschetzduality theorem and the Universal coefficient theorem (Theorems 4.6 and4.4), both of which we state without proof.
Theorem 4.4 (Universal coefficient theorem for cohomology). Let X be atopological space, let A a principal ideal domain and let N be a module overA. Then the sequence
0→Ext1A(Hk−1(X;A), N)→Hk(X;N)→HomA(Hk(X;A), N)→0 (4.3)
is a short exact sequence that splits.
Corollary 4.5. Let M be a compact topological topological manifold withboundary and let F be a field. Then Hk(M,F) ' Hk(M,F).
Proof. Take M to be a compact topological manifold. Then M is certainlyalso a topological space, so that Theorem 4.4 is applicable. Let F be a fieldand take N := F in Theorem 4.4 (this works since F is trivially a moduleover itself). Now, since (4.3) splits, we have
Hk(M ;F) ' HomF(Hk(M ;F),F)⊕ Ext1F(Hk−1(M ;F),F)
Moreover, since F is a field, we have
Ext1A(Hk−1(M ;F),F) = 0,
and sinceM is a compact topological manifold, Hk(M ;F) is finite-dimensional(we will not prove this well known fact, but for smooth manifolds it followseasily from the Hodge decomposition theorem, as we will see in Chapter 5and is thus isomorphic to its dual space, that is
HomF(Hk(M ;F),F) ' Hk(M,F).
We thus conclude that
Hk(M ;F) ' HomF(Hk(M ;F),F) ' Hk(M ;F),
which is what we set out to prove.
40
Theorem 4.6 (Poincare-Lefschetz duality theorem). Let M be a compact,oriented n-dimensional topological manifold with boundary. Then, for anyring R,
Hk(M ;R) ' Hn−k(M,∂M ;R) and Hk(M,∂M ;R) ' Hn−k(M ;R).
The Poincare-Lefschetz duality theorem is a generalization of the Poincareduality theorem, with the latter following easily from the former.
Corollary 4.7 (Poincare duality theorem). Let M be a closed topologicalmanifold and let R be a ring. Then Hk(M ;R) ' Hn−k(M ;R).
Proof. Take M to be a closed topological manifold. Since M has no boundarywe have Hk(M) ' Hk(M,∂M), and then the corollary follows from applyingTheorem 4.6 to M .
Proposition 4.8. Let M be a compact topological manifold with boundary.Then, for any field F,
Hk(M ;F) ' Hn−k(M,∂M ;F) and Hk(M,∂M ;F) ' Hn−k(M ;F).
Proof. The result is a direct consequence of Corollary 4.5 and Corollary4.6.
4.3 Alexander duality
In what lies ahead, we need to be able to compare the homology of a compactdomain in Rn with the homology of its relative complement with respect toRn. Our main tool for this is the Alexander duality theorem, which we statehere without proof.
Theorem 4.9 (Alexander duality theorem). Let X be nonempty, compactand locally contractible proper topological subspace of Sn. Then, for any ringR,
Hk(Sn \X;R) ' Hn−k−1(X;R).
Corollary 4.10. Let X be nonempty, compact and locally contractible propertopological subspace of Rn. Then, for any ring R,
Hk(Rn \X;R) 'Hn−k−1(X;R) if k = n− 1
Hn−k−1(X;R) otherwise
Proof. Since Sn ' Rn∪∞, take X ′ = X ∪∞ and note that Hi(X;R) 'Hi(X
′;R) except for i = 0, since X ′ has one more connected componendthan X. This implies that H0(X;R) ' H0(X ′;R). Applying Theorem 4.9to X ′ thus yields the desired result.
In particular, the above corollary, as well as Proposition 4.8, applies tosmooth manifolds, since any smooth n-dimensional manifold can be embeddedinto R2n. The important special case that we are concerned about is whenX = D is a compact regular domain in R3.
41
4.4 Homology in R3
Let D be a compact regular domain in R3. Elements in H0(D;R) canbe thought of as equivalence classes of points in D. Two points are thenequivalent in H0(D;R) if they lie in the same connected component of D.Similarly, elements in H1(D;R) can be thought of as equivalence classesof oriented loops in D, and elements in H2(D;R) can be thought of asequivalence classes of closed oriented surfaces in D. We adopt this point ofview in this section, in which we state some properties of singular homologyfor compact regular domains in R3. We do not aim for general statementshere; we only state what we need and leave it to the reader to look up furtherdetails in eg. [6].
Proposition 4.11. Let D be a compact regular domain in R3. Then thefollowing statements are true:
1. Every element in H1(D;R) has a representative that is a closed curvein ∂D. Similarly, every element in H1(R3 \D;R) has a representativethat is a loop in ∂D.
2. Every element in H1(D;R) is expressible as the sum two classes, one ofwhich that can be representated by a closed curve that bounds a surfacein H2(D;R) and one of which that can be representated by a closedcurve that bounds a surface in H2(R \D;R)
Proof. Since Hi(Rn;R) = 0 for i > 0, the result is given by looking at theMayer-Vietoris sequence
. . . −→ H2(Rn;R) −→ H1(∂D;R)
−→ H1(D;R)⊕H1(Rn \D;R) −→ H1(Rn;R) −→ . . .
For more details, see the proof of Proposition 5.14.
Proposition 4.12. Let D be a compact regular domain in R3. Then thefollowing statements are true:
1. dimH0(D;R) = dimH3(D, ∂D;R) = the number of connected compo-nents of D.
2. dimH1(D;R) = dimH2(D, ∂D;R) =∑i∈I
gi, where I indexes the con-
nected components of ∂D and gi denotes the genus of the i-th suchcomponent.
3. dimH2(D;R) = dimH1(D, ∂D;R) = the number of connected compo-nents of ∂D − the number of connected components of D.
Proof. Let D be as in the proposition.
42
1. Since D is a manifold, its path components and connected componentscoincide. The result then follows from Remark 4.3 and Proposition 4.8.
2. Proposition 4.11 tells us that H1(∂D;R) = H1(D;R)⊕H1(R3 \D;R)and from Corollary 4.10 we know that H1(R3 \ D;R) = H1(D;R).From Corollary 4.5 we then see that dimH1(∂D;R) = dimH2(D;R),so that dimH1(∂D;R) = 2 dimH1(D;R). But is
dimH1(∂D;R) = 2 ·∑i∈I
gi, (4.4)
and hence, by Proposition 4.8, we have
dimH1(D;R) = dimH2(D, ∂D;R) =∑i∈I
gi. (4.5)
3. This must follow from the Euler characteristic, since we already com-puted all Hi(D;R) dimensions except H2(D;R) (and all of these arezero except for i = 0, 1, 2, by e.g. Alexander duality). By definition wehave
χ(D) = dimH0(D;R)− dimH1(D;R) + dimH2(D;R),
and hence
dimH2(D;R) = χ(D)− dimH0(D;R) + dimH1(D;R).
Thus, if we knew χ(D) we would be done.
The key is that χ(Y ) = 0 for any closed 3-manifold Y, since by Poincareduality we have
χ(Y ) = dimH0(Y ;R)− dimH1(Y ;R)
+ dimH2(Y ;R)− dimH3(Y ;R),
butdimH1(Y ;R) = dimH2(Y ;R) = dimH2(Y ;R)
anddimH0(Y ;R) = dimH3(Y ;R) = dimH3(Y ;R),
so everything cancels.
To make a closed manifold from D, we simply glue D to itself (withreversed orientation) along the boundary (this is called doubling) toget
0 = χ(D) + χ(D)− χ(∂D).
Since χ(∂D) =∑i∈I
(2− 2gi), we thus have χ(D) =∑i∈I
(1− gi).
43
To see that this works, we compute
dimH2(D;R) = χ(D)− dimH0(D;R) + dimH1(D;R)
=∑i∈I
(1− gi)− (# con. comp. of D) +∑i∈I
gi
= (# con. comp. of ∂D)
− (# con. comp. of D),
and the proof is done (we just need to apply Proposition 4.8).
We also need the following theorem, which we state without proof.
Theorem 4.13. Let D be a compact regular domain in R3. Then we canpick bases Σimi=1 for H2(D, ∂D;R), Cimi=1 for H1(D;R), and C ′imi=1 forH1(R3 \D;R), such that the following properties hold true:
1. Σimi=1 consists of smooth oriented cross sectional surfaces in D suchthat Σi ⊂ ∂D, for 1 ≤ i ≤ m.
2. Cimi=1 consists of smooth oriented loops in D \ ∂D.
3. C ′imi=1 consists of smooth oriented loops in R3 \D.
4. The intersection number of Σi and Cj is δij and the linking number ofCi and Cj is δij
4.5 De Rham cohomology
Let M be a smooth manifold. A differential form η ∈ Ω∗(M) is said to beclosed if dη = 0. If there exists another form ζ such that η = dζ, then η issaid to be exact. The linearity of the exterior derivative tells us that theclosed and exact k-forms are linear subspaces of Ωk(M). Since dd = 0, allexact forms are closed (that is, im d ⊂ ker d) and we get the cochain complex
0 −→ Ω0(M)d0−→ Ω1(M)
d1−→ Ω2(M) −→ . . . ,
where di denotes the restriction of d to Ωi(M). This complex is known asthe de Rham complex. The k-th de Rham cohomology group, denoted Hk
dR,is defined as the k-th cohomology group of (4.5), i.e.
HkdR(M) :=
ker dkim dk−1
Thus, for two closed k-forms η and ζ the assertion that [η] = [ζ] in HkdR(M)
is equivalent to η − ζ being exact.
44
Example 4.14. In Rn, all closed forms are exact. This follows from thePoincare lemma, which states that every closed form is exact on a contractiblemanifold. As a result of this all the de Rham cohomology groups of Rn ofdegree greater that zero are trivial.
4.6 De Rham’s theorem
De Rham’s theorem tells us that for a smooth manifold M , the de Rhamcohomology group Hk
dR(M) is isomorphic to the singular cohomology groupHk(M ;R). We state it here without proof.
Theorem 4.15 (De Rham’s theorem). Let M be a smooth manifold. Thenthe map
I : HkdR(M) −→ Hk(M ;R)
[η] 7−→(Hk(M ;R) 3 [c] −→
∫cη
)is an isomorphism.
We state the following nice and simple consequence of Theorem 4.15 as acorollary.
Corollary 4.16. A closed manifold (i.e. compact without boundary) is notcontractible.
Proof. Let M be a compact manifold without boundary. Let g be anyRiemannian metric on M , and consider the associated volume form ωg. Asωg is a top form, it is automatically closed. However, it is not exact, as thefollowing argument shows.
Suppose, in order to derive a contradiction, that ωg is exact. Thenωg = dη for some (n− 1)-form η, and by Stokes’ theorem the volume of M is∫
Mωg =
∫Mdη =
∫∂M
η = 0,
which is a contradiction, and therefore ωg cannot be exact.Since ωg is closed but not exact, the cohomology of M is nontrivial. This,
in turn, implies that M is not contractible, since cohomology is invariantunder homotopy equivalence, and a contractible space is homotopy equivalentto a point.
45
5 Hodge theory
In this chapter we use the tools that we have developed so far to take a lookat Hodge theory. In particular, we will look at how the Hodge decompositiontheorem (Theorem 5.5) provides a decomposition of the space of k-formson a Riemannian manifold with boundary into an orthogonal direct sum ofsubspaces with respect to the L2-metric introduced in Section 2.6, and howa special case of this decomposition theorem for vector fields on compactregular domains in R3 can be obtained (Theorem 5.12). For a more in-depthtreatment of the Hodge decomposition theorem for manifolds with boundary,see Schwarz’s book [12]. A more elementary introduction in the contextof vector fields on compact domains in R3 is provided by Cantarella et al.in [4]. A survey that also introduces various applications of the Hodgedecomposition theorem is given by Bhatia et al. in [1].
We begin by familiarizing ourselves with some new notation and defini-tions that will be useful in what lies ahead.
5.1 Notation and definitions
Throughout this section, unless otherwise specified, let (M, g) be a smoothRiemannian manifold with boundary. We write Ck(M) and cCk(M) for thesubspaces of Ωk(M) of closed and co-closed k-forms, i.e.
Ck(M) := η ∈ Ωk(M) | dη = 0
cCk(M) := η ∈ Ωk(M) | d∗η = 0,
and we write Ek(M) and cEk(M) for the subspaces of exact and co-exactk-forms, i.e.
Ek(M) := d(Ωk−1(M)) = η ∈ Ωk(M) | η = dζ
cEk(M) := d∗(Ωk+1(M)) = η ∈ Ωk(M) | η = d∗ζ.
Intersections of spaces are denoted by juxtaposition of letters, so that, forexample, the space CcCk(M) = Ck(M) ∩ cCk(M) is the subspace of Ωk(M)consisting of forms that are both closed and co-closed. This notation isborrowed from Cappell et al., who use it in [5].
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We also need to consider forms that satisfy certain boundary conditions.With this in mind, we define the map
t : Γ(Λk(T ∗M)|∂M ) −→ Γ(Λk(T ∗M)|∂M )
by demanding that
tη(X1, . . . , Xk) = η(X‖1 , . . . , X
‖k) ∀ X1, . . . Xk ∈ Γ(TM |∂M ), (5.1)
where X = X⊥ +X‖ is the decomposition of the vector field X along ∂Minto its tangential and normal components. We then define the map n definedby
n : Γ(Λk(T ∗M)|∂M ) −→ Γ(Λk(T ∗M)|∂M )
nη 7−→ η|∂M − tη.
For any k-form η on M , we thus have η = tη + nη, and we say that tη isthe tangential component of η and that nη is the normal component of η.
Let ı : ∂M → M be the inclusion map of the boundary. By abuse ofnotation, tη is sometimes identified with the pullback form ı∗η. But tηlives in the space Γ(Λk(T ∗M)|∂M ) whereas ı∗η lives in the space Ωk(∂M) =Γ(Λk(T ∗∂M)), and since these are two different spaces we cannot have tηequal to ı∗η. The reason why this identification still “makes sense” is clarifiedin by the following proposition and its proof.
Proposition 5.1. Let (M, g) be a Riemannian manifold with boundary, letı : ∂M →M be the inclusion map of the boundary, and let η ∈ Ωk(M). Then
tη = 0 if and only if ı∗η = 0.
Proof. Let X1, . . . , Xk ∈ Γ(TM |∂M ). From the definition of tη, we see that(tη)(X1, . . . , Xk) = 0 whenever Xi = X⊥i for some 1 ≤ i ≤ k, and hence
(tη)(X1, . . . Xk) = (tη)(X1 −X⊥1 , . . . Xk −X⊥k ).
Therefore, we only need to consider the case Xi = X‖i for all 1 ≤ i ≤ k. Then
Xi, . . . , Xk ∈ Γ(T∂M) and we have
(tη)(X1, . . . Xk) = (ı∗η)(X1, . . . , Xk),
and we are done with the proof.
Proposition 5.2. Let (M, g) be a Riemannian manifold with boundary andlet η ∈ Ωk(M). Then
?(nη) = t(?η) and ? (tη) = n(?η). (5.2)
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Proof. Let N denote the outward-pointing unit normal vector field alon ∂Mand let E1, . . . , En be a local orthonormal frame such that E1|∂M = N andE2|∂M , . . . , En|∂M ∈ T∂M . The above definitions, together with Proposition2.17 and Remark 2.18, imply that for any (k, n)-shuffle σ ∈ S(k, n) ⊂ Sn (asdefined in Proposition 2.9), we have
(?nη)(Eσ(k+1), . . . , Eσ(n)) = (sgn σ) · (η|∂M − tη)(N,Eσ(2), . . . , Eσ(k))
= (sgn σ) · (η|∂M )(N,Eσ(2), . . . , Eσ(k))
and
(t ? η)(Eσ(k+1), . . . , Eσ(n)) = (sgn σ) · (η|∂M − nη)(N,Eσ(2), . . . , Eσ(k))
= (sgn σ) · (η|∂M )(N,Eσ(2), . . . , Eσ(k))
whenever σ(1) = 1, and we have
(?nη)(N,Eσ(k+2), . . . , Eσ(n))
= (sgn σ) · (nη)(Eσ(1), . . . , Eσ(k))
= (sgn σ) · (η|∂M − tη)(Eσ(1), . . . , Eσ(k)) = 0
and(t ? η)(N,Eσ(k+2), . . . , Eσ(n)) = 0.
whenever σ(k + 1) = N (these are the only possible cases). This proves thefirst identity. Applying this identity to ?η and taking the Hodge star of eachside shows that
? ? [n(?η)] = ?t(? ? η) =⇒ n(?η) = ?t(η),
where we have used that ?? = ±1, and the proof is done.
We denote by ΩkD(M) the space of all k-forms that satisfy the Dirichlet
boundary condition, that is
ΩkD(M) := η ∈ Ωk(M) | tη = 0. (5.3)
Similarly, we denote by ΩkN (M) the space of all k-forms that satisfy the
Neumann boundary condition, that is
ΩkN (M) := η ∈ Ωk(M) | nη = 0. (5.4)
The space ΩkN (M) can be characterized in an alternative way, as the following
proposition shows.
Proposition 5.3. Let (M, g) be a Riemannian manifold with boundary.Then Ωk
N (M) is given by
ΩkN (M) = η ∈ Ωk | t(?η) = 0.
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Proof. Let (M, g) be a Riemannian manifold with boundary. From (5.4) weknow that a k-form η is in Ωk
N (M) if and only if nη = 0, which is the case ifand only if ?(nη) = 0. But from Proposition 5.2 we know that ?(nη) = t(?η).Thus, η is in Ωk
N (M) if and only if t(?η) = 0.
Boundary conditions are applied to Ek(M) and cEk(M) as follows:
EkD(M) := d(Ωk−1D (M)) = η ∈ Ωk(M) | η = dζ, ζ ∈ Ωk−1
D (M)
EkN (M) := d(Ωk−1N (M)) = η ∈ Ωk(M) | η = dζ, ζ ∈ Ωk−1
N (M)
cEkN (M) := d∗(Ωk+1N (M)) = η ∈ Ωk(M) | η = d∗ζ, ζ ∈ Ωk+1
N (M)
cEkD(M) := d∗(Ωk+1D (M)) = η ∈ Ωk(M) | η = d∗ζ, ζ ∈ Ωk+1
D (M).
For the other subspaces the situation looks as follows:
CkD(M) = ΩkD(M) ∩ Ck(M) CkN (M) = Ωk
N (M) ∩ Ck(M)
cCkD(M) = ΩkD(M) ∩ cCk(M) cCkN (M) = Ωk
N (M) ∩ cCk(M)
CcCkD(M) = ΩkD(M) ∩ CcCk(M) CcCkN (M) = Ωk
N (M) ∩ CcCk(M).
Proposition 5.4. Let (M, g) be a Riemannian manifold with boundary.
1. The differential d : Ωk(M)→ Ωk+1(M) preserves Dirichlet boundaryconditions, whereas the codifferential d∗ : Ωk(M)→ Ωk−1(M) preservesNeumann boundary conditions.
2. The Hodge star operator ? provides the following vector space isomor-phisms:
Ck(M) ' cCn−k(M)
Ek(M) ' cEn−k(M) EkD(M) ' En−kN (M)
CcCk(M) ' CcCn−k(M) CcCkD(M) ' CcCn−kN (M).
Proof. Let (M, g) be a Riemannian manifold with boundary.
1. Let η ∈ ΩkD(M). Then tη = 0, and by Proposition 5.1 we also have
ı∗η = 0. Since d commutes with pullbacks we have d(ı∗η) = ı∗(dη) = 0,and so t(dη) = 0. That is, dη lies in Ωk+1
D (M).
Next, let η ∈ ΩkN (M). Then t(?η) = 0, and again, by Proposition 5.1,
we have ı∗ ? η = 0. Using that ?? = ±1, we compute
ı∗(?d∗η) = ±ı∗(? ? d ? η) = ±ı∗(d ? η) = 0,
It then follows from Proposition 5.2 that t(?d∗η) = 0, which, usingProposition 5.4, proves that d∗η ∈ Ωk−1
N (M).
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2. Let η ∈ Ωk(M). From the definition of the codifferential it is clear thatdη = 0⇔ d∗(?η) = 0 and that d∗η = 0⇔ d(?η) = 0. This implies that?Ck(M) = cCn−k(M) and that ?CcCk(M) ' CcCn−k(M). Takinginto account part 1 of this proposition, it is also clear that ?Ek(M) =cEn−k(M). Lastly, using Proposition 5.2 we get
?ΩkD(M) = ?η | η ∈ Ωk(M), tη = 0
= η | ? η ∈ Ωn−k(M), t(?η) = 0
= η ∈ Ωn−k(M) | nη = 0
= Ωn−kN (M).
Finally, since ?? = ±1, we can apply ? again to get isomorphisms inthe reverse directions.
5.2 The Hodge decomposition theorem
In this section, the reference to M is understood, and will therefore beomitted. We thus write Ωk instead of Ωk(M) etc. Also, throughout theremainder of the text, we will write + for direct sums, and reserve the use of⊕ for orthogonal direct sums.
Theorem 5.5 (Hodge decomposition theorem). Let (M, g) be a compact,oriented, smooth Riemannian n-manifold with boundary. The space Ωk ofsmooth differential k-forms on M decomposes as the orthogonal (with respectto the L2-inner product on k-forms) direct sum
Ωk = cEkN ⊕ CcCk ⊕ EkD. (5.5)
Moreover, CcCk decomposes as
CcCk = CcCkN ⊕ EcCk = CcEk ⊕ CcCkD. (5.6)
The Hodge decomposition theorem can be split into two statements, oneconcerning span and the other concerning orthogonality. What the spanstatement tells us is that the various subspaces that appear as summands inthe theorem actually span the space of differential k-forms, i.e. that
Ωk = span(cEkN ∪ CcCk ∪ EkD) (5.7)
andCcCk = span(CcCkN ∪ EcCk) = span(CcEk ∪ CcCkD). (5.8)
The orthogonality statement then tells us that these subspaces are orthogonal.That is, that the spaces cEkN , CcCk and EkD are mutually orthogonal, as
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are CcCkN and EcCk, and also CcEk and CcCkD. Whereas the proof ofthe span statement is somewhat involved and requires analysis, the proofof the orthogonality statement is rather straightforward. We prove theorthogonality statement here. A proof of the span statement, which relies onresults about the solvability of boundary value problems for certain partialdifferential equations, will be given only for Theorem 5.12, which is a specialcase of the Hodge decomposition theorem dealing with the decomposition ofvector fields on compact domains in R3. This proof is the subject of Chapter7.
Remark 5.6. For a closed manifold, the boundary conditions in Theorem5.5 are vacuous, and the decomposition (5.5) becomes
Ωk = cEk ⊕ CcCk ⊕ Ek.
Also, EcCk = 0 and CcEk = 0.
Proof of orthogonality statement. We need to show that the summands in(5.5) and (5.6) are orthogonal.
We begin by showing that EkD is orthogonal to cCk. Take η = dζ ∈ EkDand θ ∈ cCk. Then, using Stokes’ theorem, we compute
〈dζ, θ〉 =
∫Mdζ ∧ ?θ
=
∫Md(ζ ∧ ?θ)±
∫Mζ ∧ (d ? θ)
=
∫∂M
ζ ∧ ?θ = 0,
where the last equality holds because ζ lives in Ωk−1D , and hence the restriction
of ζ to ∂M is zero.To show that cEkN is orthogonal to Ck, take η = d∗ζ ∈ cEkN and θ ∈ Ck.
Again, using Stokes’ theorem, we see that
〈d∗ζ, θ〉 =
∫Mθ ∧ ?(d∗ζ)
= ±∫Mθ ∧ d(?ζ)
= ±∫Md(θ ∧ ?ζ) = ±
∫∂M
θ ∧ ?ζ = 0,
where the last equality holds because ζ lives in Ωk+1N and ? takes Ωk
N to Ωn−kD ,
and thus the restriction of ?ζ to ∂M is zero.Since EkD ⊂ Ck, it now follows that EkD is orthogonal to cEkN .Next, we show that EcCk is orthogonal to CcCkN . Let η = dζ ∈ EcCk
and θ ∈ CcCkN . Since d∗θ = 0, it must be the case that d ? θ = 0. Moreover,
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since ?θ lives in ΩkD, the restriction of ?θ to ∂M is zero. Using these facts,
we compute
〈dζ, θ〉 =
∫Mdζ ∧ ?θ
=
∫Md(ζ ∧ ?θ)±
∫Mβ ∧ (d ? θ)
=
∫∂M
ζ ∧ ?θ = 0,
i.e. η and γ are orthogonal.Lastly, to see that CcEk is orthogonal to CcCkD, we observe that ?
maps CcEk to EcCn−k and CcCkD to CcCn−kN , and that 〈η, ζ〉g = 〈?η, ?ζ〉whenever η, ζ ∈ Ωk (see propositions 2.23 and 5.4).
5.3 Hodge isomorphism theorem
Closely related to the Hodge decomposition theorem is the Hodge isomor-phism theorem (also known as Hodge’s theorem). We discuss it brieflyhere.
Let (M, g) be a compact, oriented Riemannian manifold without bound-ary, and consider a k-form η on M . By the Hodge decomposition theoremwe know that there exists an orthogonal decomposition of η with respect tothe L2-inner product on differential forms as
η = dα+ d∗β + γ,
where α ∈ Ωk−1(M), β ∈ Ωk+1(M) and γ ∈ CcCk(M). This decompositionis unique in the sense that dα, d∗β and γ are uniquely determined by η,although α and β are not.
A k-form ζ represents a class in HkdR(M) if it is closed. Going back to
η, we see that the closed forms in its decomposition are dα and γ, whereasd∗β is not closed, and hence not exact (unless it is the zero form). From thiswe conclude that η represents a class in Hk
dR(M) if and only if d∗β = 0, inwhich case we get
η = dα+ γ.
Moreover, it follows from the uniqueness of the Hodge decomposition thatγ is the only harmonic representative of its cohomology class. This result,which we recovered with ease from the Hodge decomposition theorem, isknown as the Hodge isomorphism theorem.
Theorem 5.7 (Hodge isomorphism theorem). Let M be a compact, orientedRiemannian manifold without boundary. Then every cohomology class inHk
dR(M) has a unique harmonic representative.
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It is well known that the kernel of an elliptic operator is finite dimensional.Since on a Riemannian manifold the Laplace-de Rham operator ∆ is anelliptic operator, we get the following corollary to Theorem 5.7.
Corollary 5.8. Let M be as in Theorem 5.7. Then the cohomology of M isfinite dimensional.
Let γ be the harmonic. Then every representative of the cohomologyclass of γ is of the form γ + dα. The norm of a such element is
||γ + dα||2 = ||γ||2 + 2〈γ, dα〉+ ||dα||2 = ||γ||2 + ||dα||2,
whence it follows that the element that minimizes the norm in a given classis its harmonic representative. We state this result as a corollary.
Corollary 5.9. Let M be as in Theorem 5.7. Then the unique element thatminimizes the norm in a given cohomology class in Hk
dR(M) is its harmonicrepresentative.
5.4 Hodge decomposition in three-space
Throughout this section, unless otherwise stated, let D be a compact regulardomain in R3. As before, Γ(TD) denotes the vector space of smooth vectorfields on D. We define an L2-inner product, denoted 〈·, ·〉, on Γ(TD) by
〈·, ·〉 : Γ(TD)× Γ(TD) −→ R
(V,W ) 7−→∫DV ·W dV.
(5.9)
A fact that will be useful is that this metric is related to the L2-metric ondifferential 1-forms via the musical isomorphisms [ and ]. The followingproposition makes this statement precise.
Proposition 5.10. Let η, ζ ∈ Ω1(D) and V,W ∈ Γ(TRn). Then
〈η, ζ〉g = 〈η], ζ]〉 and 〈V,W 〉 = 〈V [,W [〉g.
Proof. Let D be a compact regular domain in Rn with smooth boundaryand let η, ζ ∈ Ω1(D). Since by Proposition 3.6 we have ∗(η] ∧ ∗ζ]) = η] · ζ],we get
〈η, ζ〉g =
∫Dη ∧ ∗ζ
=
∫D?(η ∧ ∗ζ) ωg
=
∫Dη] · ζ] dV = 〈η], ζ]〉,
which shows that ] preserves the L2-inner product. It follows that [ preservesthe inner product too, since it is the inverse isomorphism to ].
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Remark 5.11. Note that Proposition 5.10 implies that orthogonality ispreserved by [ and ].
Using the L2-inner product on vector fields on D, we obtain a specialcase of the Hodge decomposition theorem that deals with vector fields on D.Before we state the theorem, we need to introduce a five subspaces of Γ(TD).The names of these subspaces may appear cryptic at this point, but later onwe will come up with alternative characterizations that explain our notationin Chapter 7. The notation is originally found in [4]. The subspaces are
FK(D) = (ker(∇×))⊥
HK(D) = ( im (∇))⊥ ∩ (ker(∇×))
CG(D) = ( im (∇)) ∩ ( im (∇×))
HG(D) = ( im (∇×))⊥ ∩ (ker(∇·))
GG(D) = (ker(∇·))⊥,
where ∇ is understood to be the map ∇ : C∞(D)→ Γ(TD), and ∇× and ∇·are understood to be the maps ∇× : Γ(TD)→ Γ(TD) and ∇· : Γ(TD)→ R.
We now turn our attention to the Hodge decomposition of Γ(TD). Inwhat follows, unless otherwise stated, the reference to the manifold D willbe understood and omitted. For example, we write FK instead of FK(D)when the reference to D is understood.
Theorem 5.12. Let D be a compact regular domain in R3. The space Γ(TD)of smooth vector fields on D, endowed with the L2-inner product 〈·, ·〉, canbe expressed as the orthogonal direct sum
Γ(TD) = FK ⊕HK ⊕ CG⊕HG⊕GG, (5.10)
where
im (∇) = CG⊕HG⊕GG
im (∇×) = FK ⊕HK ⊕ CG
ker(∇×) = HK ⊕ CG⊕HG⊕GG
ker(∇·) = FK ⊕HK ⊕ CG⊕HG,
The following lemma will turn out to be useful when proving Theorem5.12. Note that this lemma holds for more general manifolds than those towhich Theorem 5.12 is applicable.
Lemma 5.13. Let (M, g) be a compact, oriented, smooth Riemannian n-manifold with boundary. Then Ek(M) = EcCk(M)⊕EkD(M) and cEk(M) =CcEk(M)⊕ cEkN (M).
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Proof. First, we show that Ek(M) = EcCk(M)⊕ EkD(M).Let η ∈ Ek(M). Since η is exact we have η = dα for α ∈ Ωk−1(M). By
the Hodge decomposition theorem we have
Ωk(M) = cEkN (M)⊕ CcCkN (M)⊕ EcCk(M)⊕ EkC(M). (5.11)
It is clear that EcCk(M)⊕ EkD(M) ⊂ Ek(M). Thus, in order to prove thatEk(M) = EcCk(M)⊕EkD(M), it suffices to show that Ek(M) is orthogonalto cEkN (M) and CcCkN (M), which implies, as is immediate from (5.11), thatEk(M) ⊂ EcCk(M)⊕ EkD(M).
To this end, suppose first that ζ ∈ cEkN (M). Using Stoke’s theorem andthat d(ϕ ∧ ψ) = dϕ ∧ ψ + (−1)kϕ ∧ dψ for ϕ ∈ Ωk(M) and ψ ∈ Ω∗(M), wecompute
〈η, ζ〉 =
∫Mdα ∧ ?ζ =
∫Md(α ∧ ?ζ)− (−1)k
∫Mα ∧ ?d∗ζ
=
∫∂M
α ∧ ?ζ − (−1)k∫Mα ∧ ?d∗ζ. (5.12)
Note that since ζ ∈ ΩkN (M) we have ?ζ ∈ Ωn−k
D (M) and hence ?ζ = 0 on∂M . We also have d∗ζ = d∗d∗β = 0. Thus both integrals in (5.12) are zero,i.e. 〈η, ζ〉 = 0.
Suppose now that ζ ∈ CcCkN (M). Again we have ?ζ ∈ Ωn−kD (M) and
again, since ζ is coclosed, d∗ζ = 0, so that the integrals in (5.12) are zeroand 〈η, ζ〉 = 0. We thus conclude that Ek(M) = EcCk(M)⊕ EkD(M).
Next, we show that cEk(M) = CcEk(M) ⊕ cEkN (M). Using the resultjust obtained together with the fact that the Hodge star preserves innerproducts (Proposition 2.23) as well as provides the isomorphisms ?cEk(M) =Ek(M), ?EcCk(M) = CcEk(M) and ?EkD(M) = cEkN (M), we see thatcEk(M) = CcEk(M) ⊕ cEkN (M) emerges as the dual result to Ek(M) =EcCk(M)⊕ EkD(M) under ?.
We need another lemma in the proof of Theorem 5.12, which is an easyCorollary to the following proposition.
Proposition 5.14. Let D be a compact regular domain in Rn. Then CkD ⊂Ek.
Proof. The boundary ∂D has a bi-collar (meaning that it has a neighbour-hood in Rn that looks like ∂D×(−ε, ε), with ∂D sitting inside as ∂D×0, withnegative part U− = ∂D× (−ε, 0) in D and positive part U+ = ∂M × (0, ε) inRn\D. This lets us cover Rn by two open sets A = D∩U and B = (Rn\D)∩U ,with intersection A ∪B = U .
Applying the Meyer-Vietoris sequence for de Rham cohomology yields
HkdR(Rn) = 0 −→ Hk
dR(A) = HkdR(D)⊕Hk(B) = Hk
dR(Rn \D)
−→ HkdR(U) = Hk
dR(∂D) −→ Hk+1dR (Rn) = 0.
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When k 6= 0, we get the middle map
HkdR(D)⊕Hk
dR(Rn \D)→ HkdR(∂D)
which is (ı∗−∗) (that is, induced by inclusions), and which is an isomorphism.In particuar, ı∗ is injective. When k = 0, ı∗ sends locally constant functionson D to their restriction to ∂D. Also in this case ı∗ is injective. In morehuman terms, if a form representing class a ∈ Hk
dR(D) becomes exact afterrestricting to the boundary, then it is already exact on D.
Certainly, a form which is closed and satisfies Dirichlet boundary conditionbecomes exact (in fact zero!) when restricted to the boundary. So it mustbe exact on D.
Lemma 5.15. Let D be a compact regular domain in Rn. Then CcCkN andCcCkD are orthogonal.
Proof. By Proposition 5.14, CcCkD ⊂ CkD is exact. Applying Theorem 5.5then yields the desired result.
Remark 5.16. Theorem 5.5 does not state that CcCkN and CcCkD areorthogonal for the simple reason that, in general, they are not. The problem ofdetermining for which manifolds the are orthogonal is discussed by Shonkwilerin [13]. The particular case for domains in R3 is then dealt with by Poelke,using Shonkwiler’s results, in [10].
We are now in a position to derive the Hodge decomposition theorem forvector fields on compact regular domains in R3, i.e. 5.12, from the generalHodge decomposition theorem, i.e. 5.5.
Proof of Theorem 5.12. In this proof, we will only consider 1-forms on D,and will therefore omit superscripts in order to improve legibility. Forexample, we write E instead of E1 when referring to exact 1-forms on D.
First, suppose that η is closed 1-form in cEN . Then η is also coclosed(since d∗d∗ = 0) and hence η ∈ CcC, which contradicts (5.5) unless η = 0.Hence the only closed 1-form in cEN is the zero form, and since all forms inCcC and ED are closed, it follows from Theorem 5.5 that C = CcC ⊕ EDand that
cEN = C⊥.
For similar reasons we have cC = CcC ⊕ cEN , and hence
ED = cC⊥.
It follows that
C = CcC ⊕ ED = CcCN ⊕ EcC ⊕ ED = CcCN ⊕ E,
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where the last equality comes from Lemma 5.13, and from which it followsthat CcCN is the orthogonal complement of E in C, that is
CcCN = E⊥ ∩ C
Similarly, again using Lemma 5.13, we have
cC = CcC ⊕ cEN = CcCD ⊕ CcE ⊕ cEN = CcCD ⊕ cE,
thus CcCD is the orthogonal complement of cE in cC, that is
CcCD = cE⊥ ∩ cC
Lastly, since EcE is a subset of CcC that is orthogonal to both CcCN andCcCD, and since by Lemma 5.15 CcCN is orthogonal to CcCD, we have thedecomposition
CcC = CcCN ⊕ EcE ⊕ CcCD,
which lets us express Ω as the direct sum of five mutually orthogonal subspacesas follows:
Ω = cEN ⊕ CcC ⊕ ED= cEN ⊕ CcCN ⊕ EcE ⊕ CcCD ⊕ ED
= C⊥ ⊕ (E⊥ ∩ C)⊕ EcE ⊕ (cE⊥ ∩ cC)⊕ cC⊥.
This concludes the first part of the proof.For the second part of the proof, we need to identify the five subspaces in
the above decomposition with subspaces of Γ(TY ) via a suitable isomorphism.Recall diagram (3.9). From it we deduce that the two diagrams
C∞(R3) Γ(TR3) Γ(TR3)
Ω0(R3) Ω1(R3) Ω2(R3)
∇
id
∇×
[ β
d d
(5.13)
and
Γ(TR3) Γ(TR3) C∞(R3)
Ω2(R3) Ω1(R3) Ω0(R3)
∇×
β
−∇·
[ id
d∗ d∗
(5.14)
are commutative. Since D is a regular domain in R3, these diagrams workequally well with D substituted for R3 . This suggest that [−1 = ] is our
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isomorphism of choice. Indeed, it follows from the commutativity of thediagrams that ] provides the following isomorphisms
E∼−−−→ im (∇)
cE∼−−−→ im (∇×)
C∼−−−→ ker(∇×)
cC∼−−−→ ker(∇·),
that is to say E] = η] | η ∈ Ω = im (∇) and so forth.Now, recall that by Proposition 5.10), orthogonality is preserved by ].
From the above discussion we thus conclude that summands in (5.10) aregiven by
(cEN )] = (C⊥)] = (ker(∇×))⊥ = FK
(CcCN )] = (E⊥ ∩ C)] = (E⊥)] ∩ C] = ( im (∇))⊥ ∩ (ker(∇×)) = HK
(EcE)] = ( im (∇)) ∩ ( im (∇×)) = CG
(CcCD)] = (cE⊥ ∩ cC)] = (cE⊥)] ∩ cC] = ( im (∇×))⊥ ∩ (ker(∇·)) = HG
(ED)] = (cC⊥)] = (ker(∇·))⊥ = GG.(5.15)
Finally, since ] preserves orthogonality, these subspaces are orthogonal.
Remark 5.17. The above proof could just as well have been carried out inthe setting of 2-forms on D instead of 1-forms. Diagram (3.9) shows thatβ−1 is a suitable substitute for ] in this case. The fact that β−1 = ]? onD together with the way ? maps the various subspaces of Ω1 in the Hodgedecomposition isomorphically to those of Ω2 suggests that a proof using2-forms could be carried out more or less the same way as the proof we havepresented here.
Remark 5.18. To summarize the contents of this chapter in terms ofdifferential forms, we have the following diagram for compact regular domainsin Euclidean n-space:
Ωk = cEkN ⊕ CcCkN ⊕ EcE ⊕ CcCkD ⊕ EkD
ker(d) = CcCkN ⊕ EcE ⊕ CcCkD ⊕ EkD
im (d) = ⊕ EcE ⊕ CcCkD ⊕ EkD
im (d∗) = cEkN ⊕ CcCkN ⊕ EcE ⊕
ker(d∗) = cEkN ⊕ CcCkN ⊕ EcE ⊕ CcCkD
Since CcCkN and CcCkD are not orthogonal in general on Riemannian mani-folds with boundary, we must rely on Proposition 5.14 to obtain this five-termversion of the Hodge decomposition.
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6 The Biot-Savart formula and bound-ary value problems
In order to prove the spanning statement of Theorem 5.12 in Capter 7,we need some results concerning the solutions to Dirichlet and Neumannboundary value problems for Poisson’s equation on compact regular domainsin R3, as well as some basic facts about the Biot-Savart formula, which isan equation that lets us compute the electric field generated by an electriccurrent.
6.1 Dirichlet and Neumann problems
We need the following theorem, which we state without proof, about solutionsto boundary value problems known as Dirichlet and Neumann problems forPoisson’s equation.
Theorem 6.1. Let D be a compact regular domain in R3, let Diki=1 denotethe connected components of D, and let N be the outward-pointing unitnormal vector field on ∂D. Then, for each f ∈ C∞(D) and u ∈ C∞(∂D),the following two statements are true.
1. The boundary value problem∇2ϕ = f in Dϕ = u on ∂D,
(6.1)
known as the Dirichlet problem, has a unique solution ϕ. Moreover, ϕis smooth on D.
2. The boundary value problem∇2ϕ = f in D
∂ϕ
∂N= u on ∂D,
(6.2)
known as the Neumann problem, has a solution ϕ if and only if theintegrability condition ∫
Di
f dV =
∫∂Di
u dS
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holds for each i ∈ 1, 2, . . . , k. Solutions are unique modulo a functionthat is constant on each Di, i.e. if ϕ1 and ϕ2 both solve (6.2) thenψ = ϕ1 −ϕ2 is constant on each connected component of D. Moreover,if ϕ is a solution to (6.2), then ϕ is smooth on D.
6.2 The Biot-Savart formula
In this section we state some of the properties of the Biot-Savart formulathat we will use in Chapter 7. For more information on the Biot-Savartformula, the article [3] by Cantarella et al. is useful. A short backgroundand history is provided by Parsley in [9].
Let D be a compact regular domain in R3 and let V be a vector field onD. Consider the map BS given by
BS(V )(x) :=
∫DV (y)×∇Φ(x, y) dV (y), (6.3)
where Φ(x, y) := − 1
4π|x− y|is the fundamental solution to Laplace’s equa-
tion in R3. We call (6.3) the Biot-Savart formula. Note that since
∇Φ(x, y) = ∇(− 1
4π|x− y|
)=
x− y4π|x− y|3
,
we can expand the right-hand side of (6.3) to get
BS(V )(x) =1
4π
∫DV (y)× x− y
|x− y|3dV (y).
Proposition 6.2. Let D be a compact regular domain in R3, let N be theoutward-pointing unit normal vector field along ∂D, and let V be a smoothvector field on D. Then the following properties are true of BS:
1. BS(V ) is smooth in D and in R3 \D.
2. ∇ ·BS(V ) = 0 in D and in R3 \D.
3. The curl of BS(V ) is given by
∇x ×BS(V )(x) = V ′(x) +1
4π∇x∫D
∇y · V (y)
|x− y|3dV (y)
− 1
4π
∫∂D
V (y) ·N|x− y|3
dS(y),
where V ′(x) =
V (x) for x ∈ D
0 for x ∈ R3 \D .
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Remark 6.3. The above proposition makes clear that if we the restrict thevector field BS(V ) to D we get an operator
BS : Γ(TD)→ Γ(TD).
This operator is called the Biot-Savart operator.
Consider now an oriented loop C in R3 \D. Let I ∈ R and, for x ∈ C, letI(x) denote the vector field with constant length I along C that is tangentto and pointing in the direction of C at each point. We want to modify (6.3)so that it works in this new setting. The idea is to thicken C into a thintube of radius r that I(x) can be extended to. We then apply (6.3) and takethe limit as r goes to zero. Thus we get
BS(I)(x) :=1
4π
∫CI(y)× x− y
|x− y|3dy. (6.4)
This expression naturally shares some of the properties of (6.3).
Proposition 6.4. Let D be a compact regular domain in R3, let C be anoriented smooth loop in R3 \D and let I ∈ R with I(x) the correspondingvector field along C. Then the following properties are true of BS:
1. BS(I) is smooth in D.
2. ∇ ·BS(I) = 0 in D.
3. ∇×BS(I) = 0 in D.
4. If D is a loop in D, then∫CBS(I) · ds = I · link(C,D),
where link(C,D) denotes the linking number of C and D.
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7 Proof of spanning statement fordomains in R3
7.1 Introduction
As we have seen, the Hodge decomposition theorem for vector fields ondomains in 3-space (Theorem 5.12) follows from the Hodge decompositiontheorem for differential forms on compact Riemannian manifolds (Theorem5.5). Recall that we only proved the orthogonality statement of Theorem 5.5.That is, we proved that the sums appearing in the theorem are L2-orthogonal.A direct consequence of this is that the sums are direct. We did not provethe spanning statement, i.e. we did not prove that equations (5.7) and (5.8)hold true.
In this chapter we proceed to prove, step by step, that for a compactregular domain D in R3, we have
Γ(TD) = span(FK ∪HK ∪ CG ∪HG ∪GG),
which we refer to as the spanning statement of Theorem 5.12. Together withthe proof of the orthogonality statement of Theorem 5.5, this provides a proofof Theorem 5.12. The proof used here is, except some minor modifications,the same as the by Cantarella et al. in [4].
We remind the reader that we write + for a direct sum and reserve theuse of ⊕ for orthogonal direct sums.
7.2 Notation and definitions
We are concerned with compact regular domains in R3. Throughout the restof this chapter D will denote such a domain, unless otherwise stated. Thatis, D ⊂ R3 will denote a properly embedded submanifold of codimension 0with boundary. We let N denote the outward-pointing unit normal vectorfield along ∂D, and adopt the convention that any expression in which Nappears is understood to apply only to ∂D.
In order to to establish the spanning statement of Theorem 5.12, we willproceed by writing Γ(TD) as the span of smaller and smaller subspaces,
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eventually ending up with five such subspaces. This is done in four splittingstatements, numbered 1 through 4. We also need to show that these fivesubspaces are precisely those that appear in Theorem 5.12. This is done infive inclusion statements, numbered 1 through 5.
We begin with Splitting statement 1, which states that Γ(TD) = span(K(D)∪G(D)), where K(D) = knots and G(D) = gradients are defined as follows:
K(D) := V ∈ Γ(TD) | ∇ · V = 0, V ·N = 0
G(D) := V ∈ Γ(TD) | ∃ϕ ∈ C∞(D) s.t. V = ∇ϕ.
Splitting statement 2 then states that K(D) = span(FK ′(D)∪HK ′(D)),where FK ′(D) = fluxless knots and HK ′(D) = harmonic knots are given by
FK ′(D) := V ∈ K(D) | all interior fluxes are zero
HK ′(D) := V ∈ K(D) | ∇ × V = 0,
where the interior flux of a vector field V ∈ Γ(TD) through a smoothorientable surface Σ ⊂ M with a given orientation and with ∂Σ ⊂ ∂M , isthe real number Φ given by
Φ :=
∫ΣV ·N dS.
That all interior fluxes of V are zero simply means that Φ = 0 for any suchsurface. In the case that V is divergence free and tangent to ∂M (which,in particular, is the case for V ∈ K(D) ⊃ FK ′(D)) the value of Φ dependsonly on the homology class of Σ in H2(D, ∂D). Hence, if Σiki=1 is a set ofsmooth orientable surfaces that form a basis for H2(D, ∂D), then a vectorfield V ∈ K(D) is also in FK ′(D) if and only if the interior flux througheach Σi is zero.
Recall that for a vector field V , exactness corresponds, via the isomor-phism [, to V = ∇ϕ, coexactness to V = ∇× U , closedness to ∇× V = 0and coclosedness to ∇ · V = 0. This follows from the commutative diagram(3.9) and we extend this terminology to include vector fields in addition todifferential forms (that is, we say V ∈ Γ(TD) is closed whenever V ∈ im (∇)etc.). Boundary conditions are translated similarly, i.e. via the musicalisomorphisms, from the definitions in Chapter 5. This leads us to the firstinclusion statement.
Inclusion statement 1. HK ′ = HK.
Proof. It is clear from the above definitions and from (5.15) that HK ′(D) =HK(D), since forms in HK ′(D) are closed, coclosed and satisfy the Neumannboundary condition.
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Next up is Splitting statement 3, which states thatG(D) = span(DFG(D)∪GG′(D)), where DFG(D) = divergence free gradients and GG′(D) =grounded gradients are given by
DFG(D) := V ∈ G(D) | ∇ · V = 0
GG′(D) := V ∈ G(D) | ϕ|∂D = 0.
From these definitions we immediately get the second inclusion statement.
Inclusion statement 2. GG′ = GG.
Proof. Vector fields in GG′(D) are precisely those that are exact with Dirich-let boundary conditions, and hence GG′(D) = GG(D).
Finally, Splitting statement 4 states thatDFG = span(CG′(D)∪HG′(D)),where CG′(D) = curly gradients and HG′(D) = harmonic gradients aregiven by
CG′(D) := V ∈ DFG(D) | all boundary fluxes are 0
HG′(D) := V ∈ DFG(D) | ϕ is locally constant on ∂D.
That all boundary fluxes of a vector field V ∈ Γ(TD) are 0 means thatthe flux of V through any connected component of ∂D is zero, and thata function ϕ ∈ C∞(D) is locally constant on ∂D is equivalent to ϕ beingconstant on each connected component of ∂D.
Inclusion statement 3. HG′ = HG.
Proof. Every vector field V = ∇ϕ ∈ HG′(D) is exact, thus in particularclosed, and coclosed. Since ϕ is locally constant on ∂D, it follows thatV is normal to ∂D, and hence satisfies the Dirichlet boundary condition.Therefore HG′(D) = HG(D).
We have already seen in the first three inclusion statements thatHK ′(D) =HK(D), HG′(D) = HG(D) and GG′(D) = GG(D), so we omit the primesymbols for these subspaces and write HK(D) instead of HK ′(D) etc. Bythe end of this chapter it will become clear that FK ′(D) = FK(D) andCG′(D) = CG(D) as well. Until that point, we keep the prime symbols inplace for these two spaces to avoid confusion as to how each space is defined.
In what follows, the reference to D is understood, so we omit it and writeK instead of K(D) etc.
7.3 Knots and gradients
Splitting statement 1. The span of K ∪G is Γ(TD).
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Proof. Let V be a smooth vector field on D and define two smooth functionsf : D → R and g : ∂D → R by f := ∇ · V and g := V ·N . It then followsfrom the divergence theorem (Theorem 3.13) that∫
Di
f dV =
∫∂Di
g dS,
where Mii∈I is the set of connected components of D. Let ϕ solve theNeumann problem
∇2ϕ = f in D
∂ϕ
∂N= g on ∂D
and define two vector fields VG and VK by VG := ∇ϕ and VK := V − VG. Bycomputing
VG ·N = ∇ϕ ·N =∂ϕ
∂N= g = V ·N,
we see that VK ·N = 0 on ∂D. Also
∇ · VG = ∇2ϕ = f = ∇ · V,
hence VK is divergence free, that is ∇ · VK = 0. Thus VK lives in Kand VG, being the gradient of ϕ, lives in G. This shows that Γ(TD) =span(K ∪G).
7.4 Splitting knots
Splitting statement 2. The span of FK ′ ∪HK is K.
Before we turn to the proof of Proposition 2, we state and prove inclusionstatement four as well as a lemma.
Inclusion statement 4. FK ′ = FK.
Proof. Recall that
FK = ∇ × U | ∇ · U = 0, U ×N = 0.
That is, FK ′ consists of those forms that are closed, coclosed and satisfy theNeumann boundary condition.
We begin by showing that FK ′ ⊂ ∇× U | ∇ · U = 0, U ×N = 0. Tothis end, let V ∈ FK ′ and define B := BS(V )|D. Then, by Proposition 6.4,∇ ·B = 0 and ∇×B = V , so we just need to fix B ×N .
Let C be a closed curve on ∂D, and consider the circulation around Cof the vector field B′ := B‖|∂D, consisting of the component of B|∂D that isparallel to ∂D. That is, consider∫
CB′ · ds. (7.1)
65
Notice that substituting B for B′ in the above integral does not changeits value. Hence, by Kelvin-Stokes theorem (Theorem 3.14), if C bounds asurface Σ that lies inside D we have∫
CB′ · ds =
∫CB · ds =
∫Σ
(∇×B) ·N dS =
∫ΣV ·N dS = 0, (7.2)
since V is fluxless. If, on the other hand, Σ lies outside D we get∫CB′ · ds =
∫CB · ds =
∫ΣBS(V ) · ds =
∫Σ
(∇×BS(V )) ·N dS = 0, (7.3)
because ∇×BS(V ) = 0 outside D.Now, since ∇× BS(V )|∂D = V |∂D is tangent to ∂D, the value of (7.1)
only depends on the homology class of the curve C in H1(∂D). Also, everyhomology class in H1(D) can be expressed as the sum of two classes; onethat bounds in D and one that bounds in D. Hence, from (7.2) and (7.3)we deduce that ∫
CB′ · ds =
∫CB · ds = 0
for any closed curve C on ∂D. Thus, B′ is a conservative vector field on ∂D,and can therefore be expressed as B′ = ∇‖f , where f : ∂D → R is a scalarpotential of B′ and ∇‖f denotes the gradient of f along ∂D.
Take ϕ : ∂D → R to be the solution of the Dirichlet problem∇2ϕ = 0 in Dϕ = f in ∂D.
and define a vector field U by U := B − ∇ϕ. Then ∇ · U = 0, ∇ × U =∇ × B = V and U × N = 0. To verify the last of these three expression,consider a vector v that is tangent to ∂D and calculate
B · v = B′ · v = ∇‖f · v = ∇ϕ · v,
whence it follows that U · v = (B −∇ϕ) · v = 0.Next, we prove that ∇U | ∇ · U = 0, U ×N = 0 ⊂ FK ′. To this end,
let U ∈ Γ(TD) be such that ∇ · U = 0 and U × N = 0 and define V byV := ∇× U .
It is clear that ∇ · V = ∇ ·∇× U = 0, and since U ×N = 0 implies thatU is orthogonal to ∂D we have, for any smooth surface Σ with ∂Σ ⊂ ∂D,∫
ΣV ·N dS =
∫Σ
(∇× U) ·N dS =
∫∂ΣU · ds = 0,
which shows that all interior fluxes of V are zero.To show that V is tangent to ∂D it is enough to recall Proposition 5.4,
which states that the codifferential (in our case ∇×) preserves Neumannboundary conditions. This concludes the proof of the lemma.
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Lemma 7.1. Let g denote the genus of D and let Σigi=1 be a set of surfacesthat form a basis for H2(D, ∂D;R). Then
1. HK ' H1(D;R) ' H2(D, ∂D;R) ' Rg.
2. For each g-tuple of real numbers (Φ1,Φ2, . . . ,Φg), there exists a uniquevector field V ∈ HK such that Φi is the flux of V through Σi fori ∈ 1, 2, . . . , g.
Proof. By Proposition 4.12, we have
H1(D) ' H1(R3 \D;R) ' H2(D, ∂D;R) ' Rg.
Let Cigi=1 and C ′igi=1 be disjoint smooth loops that constitute a basis
for H1(D;R) and H1(R3 \ D;R) respectively, and let Σigi=1 be smoothorientable surfaces that constitute a basis for H2(D, ∂D;R). Additionally,these bases are to be chosen such that the intersection number of Ci with Σj
is δij and the linking number of Ci with C ′i is δij . That this can be done isguaranteed by Theorem 4.13.
For each 1 ≤ i ≤ g, let Ii ∈ R denote the magnitude of the vector fieldIi(x) along C ′i (as in Proposition 6.4), and let B denote the vector field onD given by
B(x) :=
g∑i=1
BS(Ii)(x) =1
4π
g∑i=1
∫Ci
Ii(y)× x− y|x− y|3
dy.
It then follows from Proposition 6.4 that ∇ ·B = 0, ∇×B = 0 and that∫Ci
B · ds = Ii.
Let g : ∂D → R be given by g := B ·N and take ϕ to be the solution ofthe Neumann problem ∇2ϕ = 0 in D
∂ϕ
∂N= g on ∂D.
Now, let V = V (I1, I2, . . . , Ig) := B − ∇ϕ. Then V has the followingproperties:
∇ · V = 0, ∇× V = 0, V ·N = 0,
∫Ci
V · ds = Ii. (7.4)
From this it follows that 0 6= V (I1, I2, . . . , Ig) ∈ HK whenever at least oneof the Ii is nonzero.
Now, for i ∈ 1, 2, . . . , g, let Φi denote the flux of V through Σ, i.e.
Φi :=
∫Σi
V ·Ni dS, (7.5)
67
where Ni denotes the unit normal vector field along Σi that defines itsorientation. Note that the integral (7.5) depends only on the homology classof Σi. The fluxes Φi cannot all be zero unless V = 0, since then we wouldhave V ∈ FK ∩HK = 0 as FK and HK are orthogonal (see the commentafter Inclusion statement 4). It follows that the linear map
Rg −→ Rg
(I1, I2, . . . , Ig) 7−→ (Φ1,Φ2, . . . ,Φg)
is an isomorphism.Suppose now that W ∈ HK also has flux Φi trough the surface Σi for
i ∈ 1, 2, . . . , g. Then V −W ∈ FK ∩HK = 0, which implies that V = W .It follows that the vector fields V (I1, I2, . . . , Ig) are the only harmonic knotsin Γ(TD).
We are now in a position to prove Proposition 2.
Proof of Splitting statement 2. Let V ∈ FK, let Σiki=1 be a set of smoothorientable surfaces in D that form a basis forH2(D, ∂D). For i ∈ 1, 2, . . . , k,let
Φi :=
∫Σi
V ·N dS,
that is, Φi is the flux of V through Σi.By Lemma 7.1 there is a unique vector field VHK ∈ HK such that the
flux of VHK through Σi is Φi for i ∈ 1, 2, . . . , k. Let VFK := V −VHK ∈ K.Since all interior fluxes of VFK′ are zero, it follows that VFK ∈ FK, andhence K = span(HK ∪ FK ′).
7.5 Splitting gradients
Splitting statement 3. The span of DFG ∪GG is G.
Proof. Let ϕ ∈ C∞(D) and let V := ∇ϕ be the corresponding gradientvector field. Let ϕ1 solve the Dirichlet problem
∇2ϕ1 = 0 in Dϕ1 = ϕ on ∂D
and define ϕ2 =: ϕ−ϕ1, so that V1 := ∇ϕ1 and V2 := ∇ϕ2 satisfy V = V1+V2.Then V1 ∈ DFG since ∇ · V1 = 0 and V2 ∈ GG since ϕ2|∂D = 0, and henceG = span(DFG ∪GG).
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7.6 Splitting divergence free gradients
Splitting statement 4. The span of CG′ ∪HG is DFG.
Our proof of this proposition relies on three lemmas, which we state andprove first.
Lemma 7.2. Let m denote the number of connected components of D andlet n denote the number of connected components of ∂D. Then
HG ' H2(D;R) ' H1(D, ∂D;R) ' Rn−m.
Proof. Let D1, . . . , Dm denote the connected components of D and let∂Di1, . . . , ∂Dini denote the connected components of ∂Di for each 1 ≤ i ≤ mThen n =
∑mi=1 ni is the number of connected components of ∂D.
For each boundary component ∂Dij , let cij ∈ R be a constant and letϕ ∈ C∞(D) be the solution of the Dirichlet problem
∇2ϕ = 0 in Dϕ = cij on ∂Dij .
(7.6)
For each set of constants cij this solution is unique.Now, let V = ∇ϕ. Then V is the gradient vector field of a function that
is constant of each boundary component of D, and since ∇ · V = ∇2ϕ = 0,we conclude that V ∈ HG. Moreover, every vector field HG is of the form∇ψ with ψ solves (7.6), and is hence uniquely determined by the value of ψon each boundary component of D. Thus, to every vector v ∈ Rn we canassociate a vector field ∇ϕ ∈ HG by means of the linear map
F : Rn −→ HG
v 7−→ ∇ϕ,
where ϕ is the unique solution to (7.6) with
c11 = v1, . . . , cmnm = vn. (7.7)
The map F is not an isomorphism, as can be seen by considering the identity∇ϕ = ∇(ϕ+ c), where c is a constant. Hence, if the vector v− v′ ∈ Rn (hereunderstood as an assignment of a constant to each boundary component ofD via the correspondence (7.7)) is constant on each ∂Di =
⋃nij=1 ∂Dij , then
F (v) = F (v′). If, on the other hand, v− v′ is not constant on each ∂Di, thenF (v) 6= F (v′).
Let S ⊂ Rn denote the subspace spanned by m vectors v1, . . . , vm suchthat vi is constant on ∂Di and 0 elswhere (again understood via the corre-spondence (7.7)). Then v1, . . . , vm forms a basis for S, and thus S ' Rm.Let π denote the projection onto the quotient space Rn/S. Because of the
69
universal property of quotient vector spaces, the above discussion impliesthat there exists a unique map G : Rn/S → HG such that the diagram
Rn HG
Rn/S
F
πG
(7.8)
commutes, and that G is in fact an isomorphism.Since Rn/S ' Rn/Rm ' Rn−m, the existence of G proves that HG '
Rn−m.Finally, Proposition 4.12 tells us that dimH2(D;R) = dimH1(D, ∂D;R) =
the number of connected components of ∂D minus the number of connectedcomponents of D, i.e. n−m.
Lemma 7.3. CG′ is a subset of CG.
Proof. Recall that CG′ lies in the image of ∇ and that CG = ( im (∇)) ∩( im (∇×)). To show that CG′ ⊂ CG, we thus only need to show that CG′
lies in the image of ∇×.Let V ∈ CG′. By Proposition 6.2 we have
∇x ×BS(V )(x) = V ′(x)− 1
4π
∫∂D
V (y) ·N|x− y|3
dS(y), (7.9)
where V ′(x) =
V (x) for x ∈ D
0 for x ∈ R3 \D , and where we have used the fact
that divV = 0. We want to show that V = ∇×U for some U ∈ Γ(TD), forwhich is is enough to show that the second term in (7.9) lies in im (∇×).
To this end, let A = B \ int(D), where B ⊂ R3 is a ball containing D inits interior and int(D) = D \ ∂D, and let NB denote the outward-pointingunit normal vector field on B. Then take W := ∇ψ, where ψ is a solutionto the Neumann problem
∇2ψ = 0 in A
∂ψ
∂N= −V ·N on ∂D
∂ψ
∂NB= 0 on ∂B.
In R3 \ int(A) we then have
∇x ×BS(W )(x) = − 1
4π
∫∂(R3\A)
W (y) ·N|x− y|3
dS(y)
=1
4π
∫∂D
V (y) ·N|x− y|3
dS(y).
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In particular, since D ⊂ R3 \ int(A), the above equation holds in D, and wethus have
∇× (BS(V ) +BS(W )) = V.
This implies that V ∈ im (∇×), and we are done with the proof.
Lemma 7.4. The linear maps Φij : Rn → R given by
Φij(v) =
∫∂Dij
F (v) ·N dS
provide an automorphism G ∈ Aut(Rn/S) given by
G : Rn/S −→ Rn/S
[v] 7−→ [(Φ11(v), . . . ,Φmnm(v))].
Proof. We first prove that G is well defined. Let [v] = [v′]. Then, since v−v′is constant on each ∂Di, we have F (v) = F (v′), and thus G([v]) = G([v′]).
Note that for all v ∈ Rn, we have
ni∑j=1
Φij(v) = 0 for all i ∈ 1, . . . ,m,
since F (v) ∈ HG is divergence free.Now, let v ∈ Rn be such that G([v]) = 0. Then F (v) ∈ HG and, since
all boundary fluxes are zero, F (v) ∈ CG′. But CG′ and HG are orthogonalsince CG′ ⊂ CG (this is a direct consequence of Lemma 7.3), whence itfollows that F (v) = 0, so that v must be constant on each component ofD. From this we conclude that v ∈ S. Thus [v] = 0, and hence G is anisomorphism.
With the proofs of the lemmas behind us, we are now ready to give aproof of Proposition 4.
Proof of Splitting statement 4. Let V = ∇ϕ ∈ DFG and let Φij denotethe fluxes of V through the components ∂Dij of D. Using the isomorphismprovided by Lemma (7.4), we get a vector v of constant boundary valuescorresponding to these fluxes. Since V is divergence free, we know that theflux of V through each ∂Di is zero, and hence v can be chosen so that
ni∑j=1
cij = 0 for all 1 ≤ i ≤ m,
where we have used the identifications provided by (7.7).Now, let ψ solve the Dirichlet problem
∇2ψ = 0 in Dψ = cij on ∂Dij .
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Then V1 = ∇ψ ∈ HG has the same boundary fluxes as V , and V2 =V − V1 = ∇(ϕ − ψ) ∈ CG′ since its boundary fluxes are all 0, and henceDFG = span(CG′ ∪HG).
7.7 Putting everything together
Inclusion statement 5. CG′ = CG.
Proof. From Lemma 7.3 we have CG′ ⊂ CG, and from Spanning statements1 through 4 and Inclusion statements 1 through 4 we have
Γ(TM) = span(FK ∪HK ∪ CG′ ∪HG ∪GG). (7.10)
Hence, by Theorem 5.12, CG′ must be equal CG.
We are now finally in a position to prove the spanning statement in R3.
Proof of spanning statement in R3. The result follows immediately from Split-ting statements 1 through 4 together with Inclusion statements 1 through 5.We get
Γ(TM) = span(FK ∪HK ∪ CG ∪HG ∪GG), (7.11)
which is what we set out to prove.
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