The Whitney embedding theoremMilton Persson

VT 2014Examensarbete, 15hpKandidatexamen i matematik, 180hpInstitutionen för matematik och matematisk statistik

AbstractA fundamental theorem in differential geometry is proven in this essay. It is theembedding theorem due to Hassler Whitney, which shows that the ever so generaland useful topological spaces called manifolds, can all be regarded as subspaces ofsome Euclidean space. The version of the proof given in this essay is very similar tothe original from 1944. Modern definitions are used, however, and many illustrationshave been made, wherever it helps the understanding.

SammanfattningEn fundamental sats i differentialgeometri bevisas i denna uppsats. Det är inbäddni-ngssatsen av Hassler Witney, som visar att de ofantligt generella och användbaratopologiska rummen kallade m̊angfalder, kan ses som delrum av n̊agot Euklidisktrum. Versionen av beviset given i denna uppsats är väldigt lik originalet fr̊an 1944.Moderna definitioner används dock, och m̊anga illustrationer har gjorts, varhelstdet hjälper först̊aelsen.

Contents

List of Figures 11. Introduction 32. Preliminaries 52.1. Manifolds 52.2. Immersions and embeddings 72.3. Bundles and polyhedra 83. Whitney’s first embedding theorem 134. Whitney’s second embedding theorem 275. Acknowledgements 45References 47

List of Figures

2.1 Example of a topological space that is not a differentiable manifold 62.2 A mapping on manifolds and the corresponding mapping on Euclidean

spaces. 72.3 Example of what is not the image of an immersion. 82.4 A 2-simplex with an illustration of how it is generated by its points. 112.5 2-dimensional polyhedron in R3 consisting of seven faces. 123.1 Dependency diagram for the path to the proof of Whitney’s first embedding

theorem. 133.2 Illustration for proof of Lemma 3.11 183.3 Illustration for proof of Lemma 3.12. 194.1 Dependency diagram for the path to the proof of Whitney’s second

embedding theorem. 274.2 Mappings of Theorem 4.5. 304.3 A completely regular immersion with one self-intersection. 324.4 A completely regular immersion considered in Lemma 4.12, Lemma 4.13,

and Theorem 4.14, as well as a construction used for the Whitney trick. 364.5 Defining vector fields necessary for the Whitney trick. 384.6 Modification of neighbourhoods, part of the Whitney trick. 394.7 Introducing a pair of regular self-intersections to a 1-dimensional manifold. 414.8 Undoing intersections. The purpose of the Whitney trick. 42

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1. Introduction

As every one knows an n-dimensional differentiable manifold is defined to be aset M together with a differentiable structure D such that the induced topologyis Hausdorff, second countable and such that M is locally Euclidean of dimensionn. Some, for example [9], assume only paracompactness instead of the strongerassumption of second countability. We can not proceed in this manner since we wantto apply Sard’s theorem. It should also be emphasized that not every set admits adifferentiable structure (see e.g. [8] or [3, 6]). This crucial point is something thatmost textbooks in differentiabl geometry forget to mention. For convenience weshall assume that our manifolds are C∞-smooth, even if the proofs go through forC1-smooth manifolds.

In [14], Scholz argue that the concept of manifolds can be traced back to Lagrangeand his “Mécanique analitique” published in 1788, other interpret it differently.We shall not here entangle in this fight. Instead, we refer the interested readerto [2, 4, 7, 10, 21] and the references therein. On the other hand, most agree onthat it was Hermann Weyl who in “Idee der Riemannfläche” published in 1913introduced for the first time what we today know as real manifolds.

The modern definition of manifolds is completely intrinsic, so we can not a prioriview a manifold globally as a subset of RN for some N . In other words, we can notassume that the manifold can be embedded in RN . This is one of the most crucialpoints of our definition. The main star of this essay is the American mathematicianHassler Whitney who in 1936 proved that any n-dimensional differentiable manifoldcan be embedded in R2n+1 ([18]). The proof of this theorem is included in mostintroductory textbooks on differential geometry (see e.g. [9, 17]), and it has beenconsiderably simplified over time. But the original work of Whitney is still to thisday very well written, and therefore we shall here follow his proof as presented in [1].In 1944, eight years after the first paper, Whitney proved in [19] the main aim ofthis essay. He prove the following:

Theorem 4.15. Every n-dimensional differentiable manifold can be embedded inR2n.

This theorem is sharp as it is shown by the example that the 2-dimensional pro-jective plane RP 2 can be embedded in R4, but not in R3. It is hard to grasp theimportance of Whitney’s embedding theorem and its proof. One thing that mightindicate this is that Field Medal laureates Simon Donaldson, Michael Freedman,Grigori Perelman, and Stephen Smale have all, directly or indirectly, been inspiredby Theorem 4.15 and its proof. Another thing is that it yields the foundation of awhole field of mathematics known as surgery theory. Finally, to quote the Frenchlegend Jean Alexandre Eugène Dieudonné in his book [2] (page 60) on Theorem 4.15:

“. . . the proof is long and difficult . . . ”To this day, seventy years after publication, the proof of Theorem 4.15 has not

been noticeably simplified. The complete proof of our focal point is rarely found inthe literature. Except for the original articles by Whitney it can be found in [1], aJapanese translation of Whitney’s original work that in 1993 was translated back

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into English. Another rare exposition was written by Prasolov; the case R2n+1can be found in [12], and the generalization can be found in [13]. In this essay wefollow [1], even though to grasp detail we have consulted [13] and [19].

One might wonder what the secret is behind the proof of Theorem 4.15. It iswhat is known today as the “Whitney Trick”, it even got its own Wikipedia page.In this essay it is divided into Lemma 4.12, Lemma 4.13 and Theorem 4.14, andit is only valid for dimension 5 or higher. Therefore, in the proof of Theorem 4.15for the lower dimensions we shall need to use the classification theorems for lowerdimensional manifolds.

The overview of this thesis is as follows. In §2, we give some necessary backgroundto differential geometry and fibre bundles. For further information we refer to [9,15, 16, 17]. In §3, we give the first proof of the first part of the embedding theoremR2n+1, and we end this essay in §4 by proving the general case (Theorem 4.15).

Before starting the proof of the all so mighty Whitney’s embedding theorem, andits trick, it should be pointed out that some depth of detail is ignored. The mainthing omitted is in Lemma 4.11 when we refer to [16] for the details of extending asection over |K ′| to |K| in the (m− i)-dimensional sphere bundle over |K| (see §2.3for definition).

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2. Preliminaries

Important or reoccurring notions and concepts of differential geometry are pre-sented in the three subsections that follow. Concepts specific to certain lemmasand theorems are defined when necessary.

2.1. Manifolds.

Definition 2.1. A set M is locally Euclidean of dimension n if every point p ∈Mhas a neighbourhood U such that there is a bijection φ from U into an open subsetof Rn. We call the pair (U, φ : U → Rn) a chart, U a coordinate neighbourhood, andφ a coordinate map on U .

Definition 2.2. Two charts (U ⊂ Rn, φ : U → Rn), (V ⊂ Rn, ψ : V → Rn) areC∞-compatible if the two maps

φ ◦ ψ−1 : ψ(U ∩ V )→ φ(U ∩ V ) ,ψ ◦ φ−1 : φ(U ∩ V )→ ψ(U ∩ V ) ,

are both C∞-smooth as maps Rn → Rn.

Definition 2.3. An atlas on a set M is a collection S = {(Uα, φα : U → Rn)}that satisfy the following:

(i) The coordinate neighbourhoods cover M , i.e.

M =⋃α

Uα .

(ii) Every image of the form φα(Uα ∩ Uβ) is open in Rn.(iii) They are pairwise C∞-compatible.

The neighbourhoods of the charts from an atlas S on a set M constitute a basis fora topology on M , called the topology induced by the atlas. Thus, the pair (M,S )is a locally Euclidean topological space. In [9] (Proposition 1.32, p. 14) one canfind the requirements necessary, on an atlas S , for the induced topology to beHausdorff and second countable. For us, the important part is that there exist suchrequirements.

Definition 2.4. Two atlases on a set M are said to be equivalent if their union isalso an atlas on M . An equivalence class of atlases on M is called a differentiablestructure on M . The pair (M,D) of a set M and a differentiable structure D onM is called a differentiable manifold if the topology induced by D makes (M,D)a Hausdorff, locally Euclidean, second countable topological space. We will oftenwrite M instead of (M,D), when the differentiable structure is understood.

Remark. Manifolds do not have to be connected. Manifolds are said to be compactif every component is compact. They are said to be non-compact no component iscompact. Until the very last part of Whitney’s second embedding theorem we willonly consider compact manifolds.

Let us now give some examples of differentiable manifolds, as well as sets thatcannot be made into differentiable manifolds.

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Example 2.5. Euclidean space Rn with differentiable structure represented by{(Rn, id)}, is a differentiable manifold, and so is every open U ⊂ Rn, with repre-sentatives {(U, id|U )}. Here, id : Rn → Rn is the identity map.Example 2.6. The cross given by

{(x, 0) |x ∈ (−1, 1)} ∪ {(0, y) | y ∈ (−1, 1)} ⊂ R2 ,

is not a differentiable manifold (see Figure 2.1). The reason is that there is nobijection between a neighbourhood of the intersection and an open subset of anyEuclidean space (see e.g. [17]).

Figure 2.1. Example of a topological space that cannot be madelocally Euclidean and hence is not a differentiable manifold.

An alternative definition of differentiable manifold is as a set M together with aso called maximal atlas. Given any atlas on M , a maximal atlas is formed as theunion of all equivalent atlases, which is the set of all compatible charts. This isprecisely what the differentiable structure consists of. Thus, given an atlas on M ,we obtain the same differentiable manifold whether we give it the correspondingmaximal atlas or the corresponding differentiable structure.

Given a set M , there is no guarantee that all charts on M will be compatible.The existence of incompatible charts implies the existence of different equivalenceclasses of charts. This, in turn, means that M may have more than one possible dif-ferentiable structure. For example, the unit sphere S31 has over 16 million differentdifferentiable structures (see [6]).

Of course, there is no guarantee that there exists a differentiable structure foran arbitrary set M . Even if M is a Hausdorff, locally Euclidean, second countabletopological space from the outset, its topology may not be the same as that inducedby a differentiable structure. In this essay, however, we will simply assume that thisproblem does not occur.

We shall now introduce the concept of submanifold. There are various definitionsin the literature but for this essay, there is only Definition 2.7.Definition 2.7. Let (M,D) be an n-dimensional manifold, and let A be a subsetof M . Regard Rk, 0 ≤ k ≤ n, as a subspace of Rn,

Rk = {(x1, . . . , xn) ∈ Rn |xk+1, . . . , xn = 0} .

Now assume that we can choose a representative S = {(Vj , ϕj)} of D such that foreach nonempty Vj ∩A, the restriction

ϕj |Vj∩A : Vj ∩A→ Rk ⊂ Rn

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Rnh1(U1)

h2(U2)

M1M2

U2U1

h2 ◦ f ◦ h−11

·x ·

f

·

h2

·

h−11

Figure 2.2. A mapping f of n-dimensional manifolds M1 andM2. The point h1(x) ∈ U1 ⊂ M1 is mapped to h2(f(x)) by thecomposition h2 ◦ f ◦ h−11 : h1(U1)→ R

n.

is a homeomorphism into an open subset of Rk and the chart (Vj∩A,ϕj |Vj∩A) is saidto be adapted to A. Thus, we have an atlas SA = {(Vj ∩A,ϕj |Vj∩A)} representinga differentiable structure DA. We say that (A,DA) is a submanifold of (M,D) ofdimension k.2.2. Immersions and embeddings.In this section we define immersions and embeddings and all that is necessary forthose definitions. We shall also give some examples of sets that can, and thatcannot, be the images of immersions or embeddings.Definition 2.8. Let (M1,D1) and (M2,D2) be differentiable manifolds. A mappingf : M1 →M2 is smooth at a point x ∈M1 if there exists charts (U1, h1), from someatlas in D1, and (U2, h2), from some atlas in D2, satisfying the following:

(i) The point x belongs to U1.

(ii) The set f(U1) is a subset of U2.

(iii) The map h2◦f ◦h−11 : h1(U1∩f−1(U2))→ h2(U2) is smooth (see Figure 2.2).If f is smooth at every point x ∈M1, then we say that f is smooth.Definition 2.9. A bijective mapping f of manifolds is a diffeomorphism if both fand f−1 are smooth.Definition 2.10. Let M1 and M2 be n-dimensional manifolds and f : M1 → M2a smooth map. For x in M1 choose charts (U1, h1) and (U2, h2) about x and f(x).We define the rank of f at x to be the rank of the Jacobian matrix of the map

h2 ◦ f ◦ h−11 : h1(U1 ∩ f−1(U2))→ h2(U2)

at h1(x).

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Figure 2.3. The double cone in R3 is not the image of an immersion.

Remark. The definition is independent of the choice of charts and it is sometimespossible to choose identity maps as h1 and h2 so that the rank of f becomes therank of its own Jacobian matrix.

Definition 2.11. A smooth map of manifolds f : M1 → M2 is an immersion ifthe rank of f at each point of M1 is the same as the dimension of M1. If f is bothan immersion and a homeomorphism, then it is an embedding. If there exists anembedding of M1 into M2, then we say that M1 is embedded in M2.

Example 2.12. The map f : R2 → S1 × R→ R3 defined by

f(u, v) = (v cosu, v sin u, v) ,

is smooth but not an embedding, since it intersects itself at (u, v) = (u, 0) (seeFigure 2.3). Examining f more closely, as follows, we see that it is not even animmersion. Since f : R2 → R3 is a map of R2, we can choose identity maps as thecoordinate maps in the definition of rank (Definition 2.10). Thus, its rank is therank of its Jacobian matrix

Df(u, v) =

cosu −v sin usin u v cosu1 0

.which has rank 1 < 2 whenever v = 0.

Example 2.13. The mapping α : R→ R2 defined by,

y = x− x(1 + x2) , z =1

(1 + x2)with graph in Figure 4.3 on p. 32, is an immersion but not an embedding since itis not injective, passing through (0, 1/2) twice. We shall get back to this examplein (4.2).

2.3. Bundles and polyhedra.This section consists of definitions necessary for Lemma 4.11, which will in turn beused in Lemma 4.12 and Lemma 4.13. For a full description of how the lemmas andtheorems of §4 depend upon one another, see Figure 4.1.

First we shall introduce the concept of topological transformation groups. Wethen need the theory of topological groups. We refer to [5] for the necessary back-ground.

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Definition 2.14. Let G be a topological group, and let Y be a topological space.Assume there is a continuous map η : G× Y → Y satisfying the following:

(i) For the identity element e of G, η(e, y) = y.

(ii) For all g, g′ ∈ G and y ∈ Y , η(gg′, y) = η(g, η(g′, y)).Then we say that G is a topological transformation group of Y , with respect to η,and that G acts or operates on Y .

Example 2.15. The general linear group of dimension n is the set of all non-singular n× n matrices GL(n,R) = {A ∈ Rnn | det(A) 6= 0}. The orthogonal groupof dimension n is the set of all orthogonal n×n matrices O(n) = {A ∈ Rnn |ATA =I} ⊂ GL(n,R). Both of these are topological transformation groups of Rn, theiridentity element is the identity matrix, e = I, and the continuous map η correspondsto matrix multiplication from the left.

Definition 2.16. A coordinate bundle B = {B, p,X, Y,G} is a collection of topo-logical spaces and continuous maps with structures satisfying the following:

(1) The members B and X are topological spaces, B is the total space and Xis the base space. The member p : B → X is a surjective continuous mapcalled the projection map of B.

(2) The member Y is also a topological space; Y is the fiber of B. The memberG is a topological transformation group called the structural group of B.

(3) The base X has an open covering {Vj}j∈J , and for each j ∈ J there is ahomeomorphism

φj : Vj × Y → p−1(Vj) ;

the Vj ’s are coordinate neighbourhoods and the φj ’s are coordinate func-tions.

(4) The coordinate functions satisfy the following:(i) p ◦ φj(x, y) = x, x ∈ Vj , y ∈ Y , j ∈ J .

(ii) The map φj,x : Y → p−1(x) defined by

φj,x(y) = φj(x, y), y ∈ Y

gives a homeomorphism of Y ,

φ−1j,x ◦ φi,x : Y → Y

for x ∈ Vi∩Vj , which agrees with the action of an element gji(x) of G.

(iii) Define a map gji : Vi ∩ Vj → G by

gji(x) = φ−1j,x ◦ φi,x .

Then gji is continuous; we say that the gji’s are coordinate transfor-mation or transition functions of B.

We write Yx for p−1(x); Yx is the fiber over x.

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Definition 2.17. For any two maps p : E → M and p′ : E′ → M with the sametarget space M , a map φ : E → E′ is said to be fiber-preserving if φ(p−1(x)) ⊂p′−1(x) for all x ∈M .Definition 2.18. A surjective smooth map p : B → X of manifolds is said to belocally trivial of rank r if

(i) each fiber p−1(x) has the structure of a vector space of dimension r;

(ii) for each x ∈ X, there are an open neighbourhood U of x and a fiber-preserving diffeomorphism φ : p−1(U)→ U × Rr such that for every x ∈ Uthe restriction

φ|p−1(x) : p−1(x)→ {x} × Rr

is a vector space isomorphism.Definition 2.19. A bundle B = {B, p,X, Y,G} in which Y = Rn and G =GL(n,R) acts on Y by the usual linear transformations is called an n dimensionalvector bundle. If p : B → X is locally trivial of rank r then B is said to be a vectorbundle of rank r.Definition 2.20. Given a manifold M , let p1 : M × Rr →M be the projection tothe first factor, i.e. p1(m,x) = m, where m ∈ M and x ∈ Rr. Then M × Rr → Mis a vector bundle of rank r, called the product bundle of rank r over M . The vectorspace structure on the fiber p−11 (x) = {(x, v) | v ∈ R

r} is given by(x, u) + (x, v) = (x, u+ v), k · (x, v) = (x, kv) for b ∈ R .

Definition 2.21. A vector bundle of rank r over a manifold M isomorphic over Mto the product bundle M × Rr is called a trivial bundle.

The remainder of this section consists of the definition of polyhedra and all thatis necessary for that definition. The importance of polyhedra in this essay comesfrom the fact that all differentiable manifolds are homeomorphic to a polyhedron.In other words, every differentiable manifold is triangulable.Definition 2.22. Let P0, . . . , Pn be affinely independent points in Rm, m ≥ n,which means that −−−→P0P1, . . . ,

−−−→P0Pn are linearly independent. We call the set

|P0P1 . . . Pn| ={X ∈ Rm

∣∣∣∣∣−−→OX = λ0−−→OP0 + · · ·+ λn−−→OPn,λ0 + · · ·+ λn = 1,}

an n-simplex. The n is the dimension of the simplex (see Figure 2.4).

Definition 2.23. With a subset {ik | 0 ≤ ik ≤ n}qk=0 of an index set {i}ni=0, any setof q + 1 points Pi0 , Pi1 , . . . , Piq , among the vertices P0, P1, . . . , Pn of an n-simplex,σ = |P0P1 . . . Pn|, are again linearly independent; hence, they define a q-simplex

τ = |Pi0Pi1 . . . Piq | ,

called a q-face of σ. If τ is a face of σ, then we writeτ ≺ σ or σ � τ .

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•P0

•P1

•P2

O•

λ0 = 1/5, λ1 = 3/5, λ2 = 1/5

•P0

•P1

•P2

O•

•P0

•P1

•P2

λ0 = 1/5, λ1 = 3/5, λ2 = 1/5

X

O•

|P0P1P2|

•P0

•P1

•P2

O•

1 23 4

Figure 2.4. A 2-simplex with an illustration of how it is generatedby its points. Faces of 2-dimensional simplices are either a point,a line or the simplex itself.

Definition 2.24. A finite set K of simplices is called a simplicial complex if itsatisfies the following:

(i) If σ ∈ K and σ � τ , then τ ∈ K.

(ii) If σ, τ ∈ K and σ ∩ τ 6= ∅, then σ ∩ τ ≺ σ and σ ∩ τ ≺ τ .The dimension of a simplicial complex K is the maximum value among the dimen-sions of the simplices belonging to K, and is denoted by dimK.

A subcollection of a simplicial complex, containing all faces of its own elements,is called a simplicial subcomplex or just a subcomplex

Definition 2.25. Let K be a simplicial complex of dimension n. The union of allsimplices belonging to K is a polyhedron of K denoted by |K|,

|K| =⋃σ∈K

σ ⊂ Rn .

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|P0P1P2|

•P0

• P1

•P2

•P3

•P4

•P5

•P6

•P7

O•

∣∣∣{|P0P1P2|, |P0P1P3|, |P1P2P6|, |P1P6P7|, |P1P4P7|, |P1P4P5|, |P1P3P5|}∣∣∣Figure 2.5. 2-dimensional polyhedron in R3 consisting of seven faces.

Simplices σ are polyhedra |K| of simplicial complices K consisting of all facesτ ≺ σ. In this sence, polyhedra are generalizations of simplices, see Figure 2.5.

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3. Whitney’s first embedding theorem

We have included a diagram below (Figure 3.1) which gives an overview of thissection and how we will reach the proof of Theorem 3.22.

The first embedding theorem.Theorem 3.22

Theorem 3.17Theorem 3.10

Corollary 3.8

Corollary 3.6

Lemma 3.21

Lemma 3.20

Lemma 3.18Lemma 3.13

Lemma 3.12

Lemma 3.11

Lemma 3.9

Lemma 3.5

Figure 3.1. Dependency diagram for the path to the proof ofWhitney’s first embedding theorem.

We follow the diagram in Figure 3.1 by proving the three theorems as soon as wecan, and by the order in which they are numbered.

Definition 3.1. An n-cube Cn(x, r) is the Cartesian product of n open intervals(a, b) centered at x and with length 2r, i.e. such that

a+ b2 = x, ‖a− b‖R

n = 2r .

When x = 0 we will omit it, writing Cn(r) for Cn(0, r).

Some measure theory is necessary before we can proceed. To be more specific,it is the concept of Lebesgue measure zero that is central to Lemma 3.5 and itscorollaries.

Definition 3.2. A cover of a topological space S is a collection of sets {Uα}α∈Asuch that S =

⋃α∈A Uα. If all the Uα’s are open, it is an open cover. A refinement of

a cover of a topological space is another cover of the same space such that every setin the refinement is a subset of at least one of the sets in the original cover. A coverof a topological space is said to be locally finite if every point has a neighbourhoodthat intersects only a finite number of sets from the cover.

Definition 3.3. A topological space X is said to be paracompact if every opencover of X has a refinement which is locally finite and open.

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Definition 3.4. A subset A of Rn is said to have measure zero if for an arbitraryε > 0 there exists an open cover of A,

A ⊂∞⋃i=1

Cn(xi, ri) ,

such that∞∑i=1

rni < ε .

Lemma 3.5. Let f : U → Rn be a smooth map, with U an open subset of Rn. IfA is an open subset of U with measure zero, then f(A) too has measure zero.

Proof. We take an n-cube C with C ∈ U and notice that A ∩ C must also havemeasure zero. Then, for an arbitrary ε > 0, there exists an open covering of A∩C,

A ∩ C ⊂∞⋃i=1

Cn(xi, ri) ,

such that∞∑i=1

rni < ε .

Furthermore, if

b = maxx∈C,i,j

∣∣∣∣∣(∂fi∂xj

)x

∣∣∣∣∣then by applying the mean value theorem in several variables to each component off , we get

‖f(x)− f(y)‖Rn ≤√b2 + · · ·+ b2n2 terms

‖x− y‖Rn =

= bn‖x− y‖Rn , x, y ∈ C .With this we have found that

f(Cn(xi, ri)) ⊂ Cn(f(xi), bnri) .

Thus

f(A ∩ C) ⊂ f( ∞⋃i=1

Cn(xi, ri))

=∞⋃i=1

f(Cn(xi, ri)) ⊂

⊂∞⋃i=1

Cn(f(xi), bnri) .

This means that f(A ∩ C) has measure zero, since∞∑i=1

2nbnnnrni = 2nbnnn∞∑i=1

rni < 2nbnnnε .

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Finally, A can be covered with a countable number of n-cubes such as C, so themeasure of f(A) must be zero. �

Corollary 3.6. With n < p let U be an open subset of Rn, and let f : U → Rp bea smooth map. Then f(U) has measure zero.

Proof. We may think of U × Rp−n as an open subset of Rp. Hence, we defineg : U × Rp−n → Rp by g = f ◦ p1, i.e.

g : U × Rp−n p1−→U f−→Rp ,

where p1 is the projection into the first component. As a composition of C∞ maps,g too, is a C∞ map. But U ∼= U × {0} ⊂ U × Rp−n has measure zero, and sof(U) = g(U × {0}) has measure zero in Rp. �

In Definition 3.4 we defined the concept of measure zero for a subset of Euclideanspace, but we must also define it for manifolds.

Definition 3.7. Let M be an n-dimensional manifold with associated differentiablestructure D , and let A be a subset of M . We say that A has measure zero ifϕα(Uα ∩A) ⊂ Rn has measure zero for an arbitrary chart (Uα, ϕα) of D .

Corollary 3.8. Let V and M be manifolds of dimensions n and m, with n < m.Let f : V →M be a smooth map. Then f(V ) has measure zero in M .

Lemma 3.9. Let M(p, n;R) be the set of real (p × n)-matrices. Since M(p, n;R)is bijective to Rpn, we give the usual topology of Rpn to M(p, n;R); thus M(p, n;R)becomes a differentiable manifold. Let M(p, n; k) ⊂ M(p, n;R) be the set of all(p × n)-matrices of rank k. If k ≤ min(p, n), then M(p, n; k) is a k(p + n − k)-dimensional submanifold of M(p, n;R).

Proof. Let E0 be an element of M(p, n; k). Without loss of generality we mayassume that E0 =

[A0 B0C0 D0

]with A0 ∈ GL(k,R). Then there exists an ε > 0 such

that det(A) 6= 0 if A is an element of GL(n,R) such that the absolute value of eachentry of A−A0 is less than ε.

Now let U ⊂M(p, n;R) be the set consisting of all (p× n)-matrices of the formE = [ A BC D ], where A is a (k × k)-matrix such that the absolute value of each entryof A−A0 is less than ε.

Then we have

E ∈M(p, n; k) if and only if D = CA−1B ,

because the rank of[Ik 0X Ip−k

] [A BC D

]=[

A BXA+ C XB +D

]is equal to the rank of E for an arbitrary ((p − k) × k)-matrix X. Now settingX = −CA−1 we see that the above matrix becomes[

A B0 −CA−1B +D

].

15

The rank of this matrix is k if D = CA−1B. The converse also holds, since if−CA−1B +D 6= 0, then the rank of the above matrix becomes greater than k.

Let W be an open subset of Rm, wherem = pn− dim(D) = pn− (p− k)(n− k) = k(p+ n− k) ,

W ={[

A BC 0

]∈M(p, n;R)

∣∣∣∣∣The absolute value of each entryof A−A0 is less than ε.}.

Then the correspondence [A BC 0

]7→[A BC CA−1B

]defines a diffeomorphism between W and the neighbourhood U ∩M(p, n; k) of E0in M(p, n; k). Therefore, M(p, n; k) is a k(p+n−k)-submanifold of M(p, n;R). �

Theorem 3.10. If f is a smooth map of an open subset U of Rn into Rp, withp ≥ 2n, then for an arbitrary ε > 0, there exists a (p×n)-matrix A = [aij ] satisfyingthe following:

(i) |aij | < ε.

(ii) The map g : U → Rp defined by g(x) = f(x) +Ax is an immersion.

Proof. We must show that the rank ofDg(x) = Df(x) +A

is n at every x ∈ U . We do this by showing that the set{A = B −Df(x) | rank B < n}

has measure zero in M(p, n;n). To show this we will use Corollary 3.8. Considerthe maps Fk : M(p, n; k)× U →M(p, n;R) defined by

Fk(Q, x) = Q−Df(x) .

Assume that k ≤ n−1. By Lemma 3.9 we have that the dimension of M(p, n; k)×Uis at most

k(p+ n− k) + n = (n− 1)(p+ n− (n− 1)) + n == (2n− p) + pn− 1

since k(p+ n− k) + n is monotonically increasing with respect to k. Furthermore,(2n− p) + pn− 1 ≤ pn = dim(M(p, n,R)) ,

since p ≥ 2n. Now, by Corollary 3.8 , the measure of Fk(M(p, n; k) × U) must bezero in M(p, n,R). Throughout all of this, no assumptions had to be made aboutthe size of the entries of A. Thus, there is a matrix A which satisfies (i) and whichis not contained in any of the Fk’s, k = 0, 1, . . . , n − 1, i.e. which does not haverank 0, 1, . . . , n− 1. This is the desired matrix A. �

16

In Lemma 3.12 a certain atlas is defined and proven to exist, which will beused in Lemma 3.18 and Lemma 3.21, Theorem 3.17 and Theorem 4.5. One ofthe properties of this atlas is that the neighbourhoods constitute a locally finiterefinement of an arbitrary open covering of the underlying manifold. Thus, theunderlying manifold must be paracompact. In the following, we use int A to denotethe interior of a set A.

Lemma 3.11. A locally compact topological space X with a countable basis is para-compact.

Proof. First, take a countable basis {Ui} of X with U i being compact. We proceedto construct a sequence of compact sets {Ai} such that

X =∞⋃i=1

Ai ,

and

Ai ⊂ int Ai+1 .

The Ai’s obtain these properties if we set A1 = U1 and

Ai+1 = (U1 ∪ · · · ∪ Uk) ∪ U i+1 ,

where k is the smallest natural number such that Ai ⊂ U1 ∪ · · · ∪ Uk.Now, we take an arbitrary open covering W = {Wj} of X and show that it has

a locally finite refinement, as follows. Since the sets Ai+1 \ int Ai are compact theycan be covered by finitely many of the Wj ’s. Therefore, there exist a finite numberof sets Vir such that

(i) Ai+1 \ int Ai ⊂⋃sr=1 Vir ,

(ii) Vir ⊂Wj for some j,

(iii) Vir ⊂ int(Ai+2) \Ai−1.(See Figure 3.2.) Setting Pi = {Vi1 , . . . , Vis} and P = P0 ∪ P1 ∪ . . . , we see thatP is a locally finite refinement of the arbitrary covering W of X. Thus, X isparacompact. �

Our manifolds, which have a countable basis, are locally compact, hence para-compact, and the atlas mentioned previously can be shown to exist.

Lemma 3.12. Let {Uα} be an open covering of an n-dimensional manifold M .Then M has an atlas {(Vj , hj)} satisfying the following:

(i) {(Vj , hj)} is denumerable (countably infinite).

(ii) {Vj} is a locally finite refinement of {Uα}.

(iii) hj(Vj) = Cn(3).

(iv) M =⋃jWj, where Wj = h

−1j (Cn(1)).

17

V31V32

V33

A2A3A4A5

Figure 3.2. Some of the V3r , which cover A4 \ int A3 and arecontained in int(A5) \A2.

Proof. (i), (ii) and (iii) are immediately satisfied as we construct a locally finiterefinement {Vj} of {Uα} such that hj(Vj) = Cn(3). For condition (iv) we firstselect a sequence of compact sets as in the proof of Lemma 3.11,

A1, A2, . . . ,

such that

M =∞⋃i=1

Ai ,

and

Ai ⊂ int Ai+1 .

With the additional requirement that

Ai+1 \ int Ai ⊂ ∪jh−1j (Cn(1)) including A1 ⊂ ∪jh−1j (C

n(1)) ,

the last condition (iv) is satisfied, as follows. What we want is ∪jh−1j (Cn(1)) =M = ∪iAi and since Ai ⊂ int Ai+1 we have⋃

i

Ai+1 \ int Ai =⋃i

Ai ,

see Figure 3.3, so

M =⋃i

Ai =⋃i

Ai+1 \ int Ai ⊂⋃j

h−1j (Cn(1)) .

But ⋃j

h−1j (Cn(1)) ⊂

⋃j

h−1j (Cn(3)) =

⋃j

Vj = M

18

A1A2A3

A2 \ int A1A3 \ int A2

Figure 3.3. Illustration of how ∪iAi+1 \ int Ai = ∪iAi.

soM ⊂

⋃j

h−1j (Cn(1)) ⊂M

i.e.M =

⋃j

h−1j (Cn(1)) =

⋃j

Wj .

�

If we can prove there is a mapping with some desirable properties on just aneighbourhood of some point in a manifold, then we can extend it to obtain a mapwith the same properties on the entire manifold using what is known as a cut-offfunction. We simply multiply the mapping with a function that is 1 where themapping is defined and has compact support containing that domain. In this essaythe following special case of such a function will suffice.

Lemma 3.13. There exists a smooth function ϕ : Rn → R satisfying the following:(i) ϕ(x) = 1, x ∈ Cn(1).

(ii) 0 < ϕ(x) < 1, x ∈ Cn(2) \ Cn(1).

(iii) ϕ(x) = 0, x ∈ Rn \ Cn(2).We call ϕ a cut-off function.

Proof. Define a function λ : R→ R by

λ(x) ={e−1/x, x > 0,

0, x ≤ 0.and set

ψ(x) = λ(2 + x)λ(2− x)λ(2 + x)λ(2− x) + λ(x− 1) + λ(−x− 1) .

Now define ϕ by

ϕ(x1, . . . , xn) =n∏i=1

ψ(xi) .

�

19

Definition 3.14. Let f0 : X → Y and f1 : X → Y be smooth maps. A homotopyfrom f0 to f1 is a smooth map F : X × [0, 1]→ Y such that

F (x, 0) = f0(x) ,F (x, 1) = f1(x)

for all x. If there exists such a smooth homotopy, then we say that f0 is a defor-mation of f1 or that f0 is homotopic to f1 and write f0 ' f1. If A ⊂ X is a closedsubset and if F (a, t) = f0(a) = f1(a) for all a ∈ A and all t ∈ [0, 1] then we say thatf0 is homotopic to f1 relative to A and we write f0 ' f1 (rel A). We sometimeswrite ft(x) = F (x, t) and call the {ft} a homotopy.Definition 3.15. If the elements ft of a homotopy {ft} are immersions for allt ∈ [0, 1] then we say that it is a regular homotopy and that f0 is regularly homotopicto f1 and write f0'

rf1. To be regularly homotopic relative to some closed subset

A ⊂ X is defined in the obvious way.Remark. In some texts, regular is the same as full rank. Then immersions areregular maps.

We denote the set of all strictly positive real numbers by R+.Definition 3.16. Let X be a topological space and (Y, d) be a metric space withmetric d, and let δ : X → R+ be a continuous function. We say that g : X → Y isa δ-approximation of f : X → Y if

d(f(x), g(x)) < δ(x) for all x in X.

Now, given a smooth map f : M → Rp (p ≥ 2n, n as the dimension of M), weprove the existence of an immersion as a δ-approximation of f . Later we will provethe existence of an embedding, given an immersion M → Rp (p ≥ 2n + 1). Thus,given a smooth map M → Rp (p ≥ 2n+ 1) we prove the existence of an embeddingM → Rp (p ≥ 2n + 1), i.e. Whitney’s first embedding theorem, which in turn isused in the proof of his second.Theorem 3.17. Let M be an n-dimensional manifold, and let f : M → Rp bea smooth map, with p ≥ 2n. There exists an immersion g : M → Rp which is aδ-approximation of f . In addition, if the rank of f is n on a closed subset N of M ,then we may choose g such that g is homotopic to f relative to N .Proof. Assume the rank of f is n on N , that is, the rank of f is n on some openneighbourhood U of N . The family {U,M \N} is an open cover of M . We select anatlas D0 = {(Vj , hj)} according to Lemma 3.12, which is a locally finite refinementof {U,M \ N} with hi(Vi) = Cn(3) and Wi = h−1i (Cn(1)). We next set Ui =h−1i (Cn(2)) and re-index the {(Vi, hi)} so that

i ≤ 0 if and only if Vi ⊂ U ,i > 0 if and only if Vi ⊂M \N .

Since U i is compact, we can setεi = min

x∈Uiδ(x) .

20

Now we construct the desired g by induction.Induction basis. Set f0 = f ; then the rank of f0 is n on U , and so it is n on⋃j≤0W j since W j ⊂ Vj ⊂ U for j ≤ 0.Induction hypothesis. Assume next that fk−1 : M → Rp is a smooth map having

rank n on Nk−1 =⋃j

Define g : M → Rp by g(x) = limk→∞ fk(x). This means the following. Recallthat

M =⋃i

Wi ,

and

N0 ⊂ N1 ⊂ N2 ⊂ . . . , Nk−1 =⋃j

Then it follows immediately that ϕj is a smooth function. For the sake of conve-nience we take

Vi ⊂ U if and only if j ≤ 0 .

Now we define the desired g by induction.Induction basis. First put f0 = f .Induction hypothesis. Assuming next that we have an immersion fk−1 : M → Rp

which is injective on⋃j

This means the following. We have

ϕk(x) =

ϕ ◦ hk(x) if x ∈ Vk ,0 if x ∈ V ck ,and, therefore,

fk(x) =

fk−1(x) + ϕ ◦ hk(x), if x ∈ Vk ,fk−1(x), if x ∈ V ck .Since {Vj} is locally finite, any given x ∈M lies only in a finite number of V ′ks. Forall other k, x ∈ V ck so that fk−1(x) = fk(x). Assume, therefore, that k0 = k0(x)is the largest k such that x ∈ Vk. Then fk(x) = fk0(x)(x) for all k ≥ k0(x) so thatg(x) = fk0(x)(x).

The definition of g(x) readily implies that g is a smooth map which is an im-mersion as well. Since k > 0 we also have that g|N = f |N so that Vk ∩ N = ∅. Itremains to show that g is injective. Assume now that g(x) = g(x0) and x 6= x0.Then from (3.4) we have

fk−1(x) = fk−1(x0), ϕk(x) = ϕk(x0) for all integers k > 0.

From the former equation we get f(x) = f(x0). Therefore, x and x0 do not belongto the same Vj . But from the latter equation we see that if x ∈ Vk for k > 0, thenx0 must also be in Vk, which cannot happen; hence, both x and x0 must be in U(the Vj were re-indexed this way). This is a contradiction as f is injective on U .

Finally, a close look at the definition of fk reveals that fk and fk−1 are regularlyhomotopic relative to N , and hence the same is true for f and g. �

Definition 3.19. Let M be a manifold and let f : M → Rp be a continuous map.We say that the set

L(f) =

y ∈ Rp∣∣∣∣∣ y = limn→∞ f(xn) for somesequence {xn}∞n=1 ⊂M whichhas no limit point in M .

is the limit set of f .

Lemma 3.20. If f : M → Rn is an injective immersion of the n-dimensionalmanifold M then

(i) f(M) is a closed subset of Rn if and only if L(f) ⊂ f(M), and

(ii) f is an embedding if and only if L(f) ∩ f(M) = ∅.

Proof. Let {xn}∞n=1 be a sequence in M , that does not converge in M . If f(M) isclosed then limn→∞ f(xn) ∈ f(M).

Now, assume that L(f) ⊂ f(M) and assume that y ∈ f(M) and define a sequence{xn}∞n=1 in M such that f(xn) ∈ Cn(y, 1/n). Then

f( limn→∞

xn) = limn→∞

f(xn) = y .

24

If limn→∞ xn ∈ f(M) then y ∈ f(M). Otherwise, y ∈ L(f) ⊂ f(M). Thus,f(M) = f(M).

Now for (ii), assume that {xn}∞n=1 is a sequence in M without limit point in M .It is true for all immersions f of M that L(f)∩ f(M) = ∅. To see this, assume thatL(f) ∩ f(M) 6= ∅ and that y ∈ L(f) ∩ f(M). Then

y = limn→∞

f(xn) ,

because y ∈ L(f). Furthermore, since f is continuous, we havef( limn→∞

xn) = limn→∞

f(xn) .

Thereforef( limn→∞

xn) = y

but y ∈ f(M) so this would imply that limn→∞ xn ∈M , which is a contradiction.Lastly, for the injective immersion f to be an embedding we only need its inverse

to be continuous. So, assume that its inverse f−1 : f(M) → M is not continuous.Then there exists a point x ∈ M such that f(x) ∈ L(f). But this contradictsf(M) ∩ L(f) = ∅. �

Remark. Note that for injective immersions f : M → Rn to be embeddings withclosed images f(M) ⊂ Rn, we need both L(f) ⊂ f(M) and L(f)∩f(M) = ∅, whichis only possible if L(f) = ∅. Hence, Lemma 3.21.

Lemma 3.21. Let M be an n-dimensional manifold. Then there exists a smoothmap f : M → Rp with L(f) = ∅.

Proof. Let {(Vj , hj)} be the atlas of Lemma 3.12 for M , and let ϕ be the smoothfunction of Lemma 3.13. For each j > 0 let ϕj : M → R be the smooth functiongiven in the proof of Lemma 3.18, equation (3.2). Set

f(x) =∑j>0

jϕj(x) .

The right-hand side of this equation makes sense because the {Vj} is locally finite.It follows that f is continuous.

We want to show that L(f) = ∅. Let {xn}∞n=1 be a sequence in M , which hasno limit point. Then for any integer m > 0 there exists an integer n > 0 withxn /∈ W 1 ∪ · · · ∪Wm, and so xn ∈ W j for some j > m. Therefore, f(xn) > m andthe sequence {f(xn)}∞n=1 has no limit point. �

Observe that we have only showed the existence of a smooth map with emptylimit set, and not an injective immersion with empty limit set. It is for thisreason alone that the immersion of Theorem 3.17 and the injective immersion ofLemma 3.18 are approximations of the given smooth map and immersion, respec-tively. That way, the injective immersion is an approximation of the smooth mapand so has empty limit set too. To see this, take a smooth function g : M → R2n+1with L(g) = ∅ and an injective immersion f : M → R2n+1 that is a δ-approximation

25

of g. Assume that {xk}∞k=1 is a sequence in M without limit point in M . Assumethat y = limk→∞ f(xk) exists. There is an ε > 0 and a k ≥ N such that

‖g(xk)− y‖R2n+1 > ε

for all N ≥ 1. Then, we can choose the continuous map δ : M → R+ as a constantmap δ = µ < ε, which would mean that limk→∞ f(xk) 6= y.

We now have all that is required for the proof of the first embedding theorem.

Theorem 3.22 (Whitney’s first embedding theorem). Every n-dimensional differ-entiable manifold can be embedded in R2n+1 as a closed subset.

Proof. Let M be an n-dimensional manifold. Let f : M → R2n+1 be a smoothmap with L(f) = ∅. Let δ : M → R+ be the constant map δ(x) = ε > 0. Then byTheorem 3.17 there exists an immersion g : M → R2n+1 which is a δ-approximationof f . Furthermore, by Lemma 3.18 there exists an injective immersion h : M →R2n+1 which is a δ-approximation of g. By the argument proceeding the proof,there is ε > 0 so small that L(h) = ∅. Hence, Lemma 3.20 implies that h is anembedding, and so h(M) is a closed subset of R2n+1. �

26

4. Whitney’s second embedding theorem

In this section we prove the main theorem of this essay, Whitney’s second embeddingtheorem, Theorem 4.15. Just like for the first embedding theorem, we have madea diagram illustrating the role of each lemma and theorem prior to Theorem 4.15(Figure 4.1).

Theorem 4.15:The secondembeddingtheorem.

Theorem 4.14

Theorem 4.9

Theorem 4.5

Theorem 3.22

Theorem 3.17

Lemma 3.12

Lemma 4.12

Lemma 4.11

Lemma 4.13

Figure 4.1. Dependency diagram for the path to the proof ofWhitney’s second embedding theorem.

We begin by defining tangent vector, tangent space and tangent bundle.

Definition 4.1. For an n-dimensional manifold M , a tangent vector vp at p ∈ Mis a linear map vp : C∞(M)→ R with the property that for f, g ∈ C∞(M),

vp(fg) = g(p)vp(f) + f(p)vp(g) .

The tangent space TpM at p ∈M is the set of all tangent vectors vp at p ∈M . Thetangent bundle of M is the disjoint union of all tangent spaces:

TM =⋃p∈M

TpM .

Definition 4.2. Let M and V be manifolds of dimensions n and p, and let f :M → V be a smooth map. A point y in V is a regular value of f if the rank of fat each point x in f−1(y) is p; otherwise, y is a critical value.

27

Definition 4.3. Let M be an n-dimensional manifold and let f : M → R2n be animmersion. We say that f is completely regular if it satisfies the following:

(i) f has no triple points, i.e. no points

f(p1) = f(p2) = f(p3) ,

with

p1 6= p2 6= p3 .

(ii) For p1 and p2, with p1 6= p2 and f(p1) = f(p2) = q,

(df)p1(Tp1M

)⊕ (df)p2

(Tp2M

)= TqR2n , (4.1)

where the differentials are defined as

(df)p(v) = Df(p) · v, v ∈M .

We say that f intersects transversely at q or that q is a regular self-intersection off , if f satisfies (ii).

We shall immediately rewrite (4.1) into a form better suited for explicit calcula-tions, by which we shall prove the transversality of a certain parametrization (4.3).The direct sum of a finite number of vector spaces coincides with the Cartesianproduct. Thus,

(df)p1(Tp1M)⊕ (df)p2(Tp2M) =

={(Df(p1) · v1, Df(p2) · v2) | v1 ∈ Tp1M,v2 ∈ Tp2M} =

={[

Df(p1) Df(p2)](v1, v2)T

∣∣∣∣∣ (v1, v2) ∈ Tp1M × Tp2M}

.

Since TqR2n = R2n for all q ∈ R2n, equation (4.1) becomes{[Df(p1) Df(p2)

]v

∣∣∣∣∣ v ∈ Tp1M × Tp2M}

= R2n .

Remark. With these rewritings we see that the dimension of the image must betwice that of the domain for there to exist regular self-intersections: For the columnvectors of the above matrix to span R2n there must be 2n of them (n for eachJacobian matrix) and they must have 2n components. The only other way for amapping to be completely regular is for it to have no self-intersections.

We prove the second embedding theorem by first proving the existence of acompletely regular immersion of an n-dimensional manifold M to R2n, Theorem 4.5.In Theorem 4.9, we show how to modify a completely regular immersion so asto introduce new regular self-intersections. With this we can ensure that thereare just as many regular self-intersections of positive type as of negative type (seeDefinition 4.8). Then, with Theorem 4.14, we can remove pairs of regular self-intersections of opposite types. Thus, we will obtain a completely regular immersion

28

with no self-intersections, i.e. an injective immersion which is the same as anembedding M → R2n when M is compact.

To prove the existance of a completely regular immersion we shall use Sard’stheorem which we state here as Theorem 4.4. A proof can be found in [11].

Theorem 4.4. Let f : U → Rp be a smooth map, defined on an open set U ⊂ Rn,and let

C = {x ∈ U | rank dfx < p} .

Then the image f(C) has measure zero in Rp.

Theorem 4.5. Let M be an n-dimensional manifold, and let f : M → R2n bea smooth map. There exists a completely regular immersion g : M → R2n whichis a δ-approximation of f . Furthermore, if f is an immersion that is compeletelyregular on some open neighbourhood U of a compact set N , then g may be chosento be regularly homotopic to f relative to N .

Proof. By Theorem 3.17 there exists an immersion f̄ which is a δ/2-approximationof f with f̄ |N = f |N . The map f̄ is completely regular on some open neighbourhoodof N . Now, select an atlas of R2n, S ′ = {(C2n(xi, 1), ψi)|i = 1, 2, . . . }, whereψi : C2n(xi, 1)→ C2n(1). The family

U = {U, (M \N) ∩ f̄−1(C2n(xi, 1)) | i = 1, 2, . . . }

is an open covering of M . We take an atlas {(Vj , hj)} of M given by Lemma 3.12with respect to U , such that for each Vj the following conditions hold.

(i) f̄ |Vj : Vj → R2n is an embedding.

(ii) For some λj such that f(Vj) ⊂ C2n(xλj , 1), there exists a diffeomorphism

ϕj : C2n(1)→ R2n

with

ϕj ◦ ψλj ◦ f̄(Vj) ⊂ C2n(1) ∩ Rn .

(We think of Rn as Rn = {(x1, . . . , xn, 0, . . . , 0) ∈ R2n} ⊂ R2n.) (SeeFigure 4.2.)

Re-indexing the {(Vi, hi)}’s, as we did in the proof of Theorem 3.17, we assumei ≤ 0 if and only if Vi ⊂ U ,

i > 0 if and only if Vi ⊂M \N .We write ϕ′j = ϕj ◦ ψλj .

We shall construct g inductively.Induction basis. Set g0 = f̄ .Induction hypothesis. Assume that we have defined gj : M → R2n with gj |N =

f̄ |N .(a) Replacing the ϕj we may assume that gj satisfies the conditions (i) and (ii)

in place of f̄ .

29

Rn

Mn

R2n

O·C2n(1)

Vj hj

C2n(3)

N

U·xλj

C2n(xλj , 1)

ψλjf̄(Vj)

f̄j

ϕj ◦ ψλj ◦ f̄(Vj) ⊂ C2n(1) ∩ Rn

Figure 4.2. Mappings of Theorem 4.5.

(b) We assume that if Nj = N ∪i≤j W i, then for any point p of gj(Nj) the set(gj |Nj )−1(p) contains at most two points, since g|N is completely regular,and in case it contains two points gj |Nj intersects transversely at p.

Induction step. Now we construct gj+1 : M → R2n. Consider the map

ϕ′j+1 ◦ gj ◦ (hj+1)−1 : Cn(3)→ C2n(1) .

Define a projection π : R2n → Rn by π(x1, . . . , x2n) = (xn+1, . . . , x2n). Then bythe choice of ϕ′j+1 we have

ϕ′j+1 ◦ gj ◦ (hj+1)−1(Cn(3)) ⊂ π−1(0, . . . , 0) = {(x1, . . . , xn, 0, . . . , 0) ∈ R2n} .

And by Sard’s theorem the set of critical values of the smooth map

π ◦ ϕ′j+1 ◦ gj : (U ∪i≤j Wi) ∩ g−1j (Uλj )→ Cn(1) ⊂ Rn

has measure zero. Hence, we may select a point cq ∈ Rn satisfying the following:(1) cq is sufficiently close to (0, . . . , 0).

(2) cq is a regular value of π ◦ ϕ′j+1 ◦ gj .

(3) If the points p1 and p2, p1 6= p2, satisfy gj(p1) = gj(p2) ∈ C2n(xλj , 1), then

ϕ′j+1 ◦ gj(p1) = ϕ′j+1 ◦ gj(p2) /∈ π−1(cq) .

30

Using the cq, we define a smooth map gj+1 : M → R2n by

gj+1(x) =

gj(x) if x ∈M \ h−1j+1(Cn(2)) ,

(ϕ′j+1)−1{ϕ′j+1 ◦ gj(x) + cqϕ(|hj+1(x)|)} if x ∈ Vj ,

where ϕ is our cut-off function. Then the conditions (1), (2), (3) for cq guaranteethat gj+1 is an immersion and that gj+1 is a δ/2q+1-approximation of gj . Further-more setting N ′j+1 = N ′j ∪W j and noting that N ′j+1 is compact, we see that for anypoint p of gj+1(N ′j+1), the set (gj+1|N ′j+1)

−1(p) contains at most two points, andif it contains two points, then gj+1 intersects transversely at p. It is also routinethat gj+1|N = f̄ |N . In addition, by readjusting the ϕj we see that gj+1 replacing f̄satisfies the conditions (i) and (ii).

Finally, by examining the definition of gj+1 carefully we see that gj+1 and gj areregularly homotopic relative to N .

Defining a map g : M → R2n byg(x) = lim

i→∞gi(x) ,

we have what we wanted. �

In order to introduce additional regular self-intersections to a given mappingM → R2n we need to construct a model of an immersion with a single regularself-intersection. The map α : R→ R2 defined by

y = x− x(1 + x2) , z =1

(1 + x2) (4.2)

is a completely regular immersion, with one regular self-intersection at (0, 1/2), seeFigure 4.3. Furthermore, there exists r > 0 so large that α is the identity mapoutside the unity disk D(0, r),

y = x− x(1 + x2) −→|x|→∞x ,

z = 1(1 + x2) −→|x|→∞0

so that, for large enough x, we have approximately

xα7→ (x, 0) .

For n ≥ 2, α : Rn → R2n is defined as follows.

α(x1, . . . , xn) = (y1, . . . , y2n), (x1, . . . , xn) ∈ Rn , (4.3)

with

y1 = x1 −2x1u, yi = xi, i = 2, . . . , n ,

and

yn+1 =1u, yn+i =

x1xiu

, i = 2, . . . , n ,

31

Figure 4.3. The completely regular immersion α : R1 → R2 de-fined by (4.2) with a single (regular) self-intersection at (0, 1/2).

where

u = (1 + x21) · · · (1 + x2n) .

The rank of a mapping is independent of the choice of coordinate maps for thecharts in the definition of rank. Since we are mapping Rn into R2n we can choosethe identity maps as coordinate maps. With this, the rank of α is simply the rankof its Jacobian matrix (Dα)(x) which is

1− 2(1−x21)

u(1+x21)4x1x2u(1+x22)

. . . 4x1xnu(1+x2n)0 1 0 00 0 . . . 0...

...0 0 . . . 1−2x1

u(1+x21)−2x2

u(1+x22). . . −2xnu(1+x2n)

x2(1−x21)u(1+x21)

x1(1−x22)u(1+x22)

. . . −2x1x2xnu(1+x2n)...

......

xn(1−x21)u(1+x21)

−2x1x2xnu(1+x22)

. . .x1(1−x2n)u(1+x2n)

.

Not all elements of the first column are zero. If x1 6= 0 then −2x1/[u(1 + x21)] 6= 0.If x1 = 0 then the first element is 1 − 2/u. If this element is zero then u = 2which means that some of the xi’s, i = 2, . . . , n, must be nonzero. Then thecorresponding term xi(1− x21)/[u(1 + x21)] = xi/u is nonzero. Thus, with the 1’s inthe other columns, there are n linearly independent nonzero columns, i.e. the rankof D(α : Rn → R2n)(x) is n and α is an immersion.

We now want to show that α has exactly one (regular) self-intersection, i.e. adouble point which intersects transversely. Let

x = (x1, . . . , xn), x′ = (x′1, . . . , x′n) ,u′ = (1 + (x′1)2) · · · (1 + (x′n)2) ,

α(x) = (y1, . . . , y2n), α(x′) = (y′1, . . . , y′2n) .

At a double point α(x) = α(x′) we have x 6= x′. First we observe that xi = x′i fori = 2, . . . , n by the definition of α. That x 6= x′ then implies that x1 6= x′1. Secondly,

32

since yn+1 = y′n+1 we must have u = u′. Combining this with the first observationwe have that 1 + x21 = 1 + (x′1)2, which is only possible if x1 = −x′1 6= 0. With this,yn+i = y′n+i can only be true if xi = 0, i = 2, . . . , n. Therefore, u = 1 + x21 andy1 = y′1 tells us that

x1 −2x1u

= −x1 +2x1u

so

1− 2u

= −1 + 2u

thus

u = 2 = 1 + x21or

x1 = ±1 .

The only double point turns out to bey = α(±1, 0, . . . , 0) .

Set x+ = (1, 0, . . . , 0) and x− = (−1, 0, . . . , 0). For α to intersect transversely weneed, by the discussion preceding the definition 4.3,{[

Dα(x+) Dα(x−)]v

∣∣∣∣∣ v ∈ Tx+Rn × Tx−Rn}

= {since TxRm = Rm, x ∈ Rm} =

={[

Dα(x+) Dα(x−)]v

∣∣∣∣∣ v ∈ R2n}

= R2n .

This is equivalent to the matrix [Df(p1) Df(p2) ] being invertible,

[Dα(x+) Dα(x−)

]=

1 0 . . . 0 1 0 . . . 00 1 . . . 0 0 1 . . . 0...

. . ....

.... . .

...0 0 . . . 1 0 0 . . . 1−1/2 0 . . . 0 1/2 0 . . . 0

0 1/2 . . . 0 0 −1/2 . . . 0...

. . ....

.... . .

...0 0 . . . 1/2 0 0 . . . −1/2

which we see is true since the columns are all linearly independent.

We shall now classify all regular self-intersections. To do that we first need somedefinitions regarding the notion of orientation.

Definition 4.6. Consider an n-dimensional manifold (M,D). Let D = [S ], S ={(Vj , ϕj)}. For x ∈ Vi ∩ Vj let aji(x) be the Jacobian matrix of ϕj ◦ ϕ−1i at ϕi(x):

aji(x) = D(ϕj ◦ ϕ−1i )(ϕi(x)), x ∈ Vi ∩ Vj .

33

Then it is easy to see thatakj(x) · aji(x) = aki(x), x ∈ Vi ∩ Vj ∩ Vk .

If we set k = i, it follows that aji(x) has an inverse. Hence aji(x) ∈ GL(n,R) andwe have a continuous map

aij : Vi ∩ Vj → GL(n,R) .

An atlas S = {(Vj , ϕj)} is oriented if for all i, j and all x ∈ Vi∩Vj , the determinant|aij(x)| is positive.

A manifold is orientable if and only if it has an orientable atlas (see [17]). We takethis to be the definition of orientable manifold. For the special case of Euclideanspaces, which are all orientable, we also need the following.

Definition 4.7. Two ordered set of n vectors, B and B′, that are bases for Rn,are said to be equivalent if their transition matrix has positive determinant. Thispartitions the set of all bases for Rn into two equivalence classes. If the transitionmatrix from a basis B of Rn to the standard basis has positive determinant, thenB is said to define the positive orientation on Rn. Otherwise it is said to define thenegative orientation on Rn.

Definition 4.8. Let f : M → R2n be a completely regular immersion of an n-dimensional manifold M .

(i) The case where M is orientable and n is even. Choose an orientation in M .Assume that f(p) = f(q), p 6= q. Let u1, . . . , un ∈ TpM and v1, . . . , vn ∈ TqM beordered sets of linearly independent vectors, which define the orientations of TpMand TqM . We say that the self-intersection at f(p) has positive type or negativetype depending on whether the ordered set of vectors

(df)p(u1), . . . , (df)p(un), (df)q(v1), . . . , (df)q(vn) ∈ Tf(p)R2n

defines the positive orientation or the negative orientation in R2n, , and we definethe intersection number of this self-intersection to be +1 or −1 accordingly. Theintersection number If of f is the sum of intersection numbers of self-intersectionsof f : If ∈ Z.

(ii) The case where M is non-orientable or n is odd. In this case we define theintersection number If ∈ Z, in the same way as above.

Theorem 4.9. Let M be an n dimensional compact manifold.(i) If M is orientable and n is even, then for an arbitrary integer m there exists

a completely regular immersion f : M → R2n with If = m.

(ii) If M is non-orientable or n is odd, then for any m ∈ Z2, there exists acompletely regular immersion f : M → R2n with If = m.

Proof. By Theorem 4.5 there exists a completely regular immersion f0 of M in R2n.Take a point x0 of M and some neighbourhood U of x0 to replace f0 by the mapβ with exactly one self-intersection defined in (4.3) or by the composite of β withthe map r : R2n → R2n defined by r(x1, . . . , x2n) = (−x1, x2, . . . , x2n). This can be

34

done in such a way that β(U)∩f0(M \U) = ∅, otherwise more self-intersections willbe introduced, including a triple point or even a quadruple point. The intersectionnumber of the new completely regular immersion f is If0 + 1 or If0 − 1. We repeatthis process till we get an immersion with intersection number m. �

Definition 4.10. Let B = {B, p,X, Y,G} be a coordinate bundle. By a section ofB we mean a continuous map s : X → B with p ◦ s = idX , the identity on X. Asmooth map s which is a section of a smooth coordinate bundle is called a smoothsection.

Lemma 4.11. Let B = {B, p, |K|,Rm, O(m)} be an m-vector bundle over a poly-hedron |K|. Let K ′ be a subcomplex of K. Assume that (i) ζ1, . . . , ζi−1 are sec-tions over |K| such that for each p ∈ |K|, ζ1(p), . . . , ζi−1(p) form an orthonor-mal system, and (ii) ζi is a section over |K ′| such that for each point p ∈ |K ′|,ζ1(p), . . . , ζi−1(p), ζi(p) form an orthonormal system. Then if dimK ≤ m − i, wecan extend ζi over |K| in such a way that the ζ1(p), . . . , ζi−1(p), ζi(p) is an orthonor-mal system.

Proof. We give only part of the proof and refer to Steenrod [16] for the rest togetherwith the necessary background in cohomology theory.

Fix a point p of |K| and let Rmp denote the fiber over p. Then the desired ζi(p)may be a unit vector in the orthogonal complement of the (i−1)-subspace spannedby ζ1(p), . . . , ζi−1(p), i.e., we want ζi(p) ∈ Sm−ip ⊂ {{ζ1(p), . . . , ζi−1(p)}}⊥ ⊂Rmp , where Sm−ip is the unit sphere in the orthogonal complement of the space{{ζ1(p), . . . , ζi−1(p)}} spanned by ζ1(p), . . . , ζi−1(p). Hence our problem reducesto the problem of extending a cross section over |K ′| to |K| in the (m − i)-spherebundle1 over |K|. �

For Lemma 4.12 and Lemma 4.13, consider a completely regular immersion f :M → R2n (see Figure 4.4 but keep in mind that the Whitney Trick is really onlyvalid for manifolds M of dimension 3 or higher). Assume f has two regular self-intersection points q and q′ of opposite sign,

f(p1) = f(p2) = q, p1 6= p2 ;f(p′1) = f(p′2) = q′, p′1 6= p′2 .

Let C1 and C2 be disjoint curves in M ; Ci connecting pi and p′i without passingthrough any other point where f has a self-intersection, i = 1, 2. Let Bi be thecurve f(Ci) connecting q to q′. Then B = B1∪B2 is a simple closed curve in f(M).

Lemma 4.12. Let A1 be the intervalA1 = {(x, 0) |x ∈ [0, 1]} ,

and A2 be the part of the circle of radius 1, centered at (1/2,−√

3/2), connectingthe endpoints r = (0, 0) and r′ = (1, 0) of A1 in the first quadrant,

A2 = {(cos(θ) + 1/2, sin(θ)−√

3/2) | θ ∈ [π/3, 2π/3]} .

1A coordinate bundle with fiber Sm−i and group O(m − i + 1).

35

f(M2)

f(M1)

q

q′

B2 = f(C2)

B1 = f(C1)

f(Mn)

p1

p′1C1

M1p′2

p2C2

M2Mn

f

ψ

x1

x2

ψ(τ ′′) = σ2

τ ′′

A1

A2

·r ·r′

τ ′

Figure 4.4. (Upper two, with the mapping f .) The completelyregular immersion f : Mn → R2n considered in Lemma 4.12,Lemma 4.13, and Theorem 4.14. (Lower two, with the mappingψ.) Construction in R2 for the Whitney trick

(See Figure 4.4.) Let A = A1 ∪A2. Let τ ′ be a small neighbourhood of A in R2 andlet τ ′′ be the region enclosed by A. Lastly, set τ = τ ′ ∪ τ ′′. With this there is anembedding ψ : τ → R2n satisfying the following:

(I) ψ(r) = q, ψ(r′) = q′ and ψ(Ai) = Bi, i = 1, 2,

(II) ψ(τ) ∩ f(M) = B,

(III) Tq∗ψ(τ) 6⊂ Tq∗f(M), where q∗ ∈ B.

Proof. We prove the theorem for an embedding ψ : τ ′ → R2n to begin with.

36

Change f slightly (by so little that it is still an embedding) at pi in Mi so thatf maps a neighbourhood of pi in Mi surjectively on the exponential image of someneighbourhood of q in

Tni := Tqf(Mi) .

Near q we assume that f maps a neighbourhood of Ci in Mi onto the exponentialimage of some neighbourhood of a line in Tni , given by its intersection with

T 2 := T1f(M1)× Tqf(M2) .

Call the lines T 1i , i.e.

T 1i := T 2 ∩ Tqf(M) .

Now we define the linear mapping φ : V̄ → T 2 which maps a closed neighbourhoodV̄ of r in τ ′ to a closed neighbourhood of q in T 2. Then the image of the restrictionφ|V̄ ∩Ai is a closed neighbourhood of q in T

1i . In a similar way we define the linear

mapping φ′ of a closed neighbourhood V̄ ′ of r′.Now we parametrize Ai,

ri : [0, 1]→ Ai ∈ R2 ,

so that φ(ri(t)) = qi(t), wherever φ is defined, meaning that these qi(t) are pointson the part of Ci mapped onto the exponential image of T 1i . Below we will, amongother things, extend V̄ and V̄ ′ over all of A in order for φ to fulfill the rest of (I)together with φ′.

Let ui(t), t ∈ [0, 1], be the vector field of tangents to Ai in τ ′, i.e.

ui(t) ∈ T (τ)|Ai ,

which satisfies the following conditions.(i)

exp(ui(t)) ⊂ τ, i = 1, 2. (4.4)

(ii) {u1(0) ∈ TrA2, u1(0) is pointed forward along A2,u1(1) ∈ Tr′A2, u1(1) is pointed backward along A2,{u2(0) ∈ TrA1, u2(0) is pointed forward along A1,u2(1) ∈ Tr′A1, u2(1) is pointed backward along A1.

(4.5)

(iii) ui(t) ∈ Tri(t)τ, ui(t) 6∈ Tri(t)Ai, i = 1, 2.

(iv)u1(t) rotates counterclockwise as t goes from 0 to 1;u2(t) rotates clockwise as t goes from 0 to 1.

(4.6)

(See Figure 4.5.)

37

x1

x2

A1

A2

•r

•r′

u2(t)

u1(t)

Figure 4.5. Defining vector fields on A = A1 ∪A2.

Currently, φ(V̄ ) lies in T 2 and contains part of T 1i which constitutes part off(Ci) = Bi. Further away from q, Bi does not, necessarily, coincide with T 1i . Sincewe want φ to satisfy (I), we extend V̄ in such a way that even if qi(t) 6∈ T 1i we stillhave qi(t) ∈ φ(V̄ ). This is done as follows. Let Ri(t) be a line segment collinearwith ui(t) and centered at ri(t). Assume that part of the boundaries of both V̄ andV̄ ′ are formed by the Ri(t)’s, whose length ρ is such that no R1(t) touches an R2(t)if r1(t), r2(t) 6∈ V̄ ∪ V̄ ′. (See Figure 4.6.)

Now define the vector field vi(t) = (dφ)ri(t)(ui(t)) for ri(t) ∈ [V̄ ∪ V̄ ′] ∪ Ai. Wehave that vi(t) is in f(V̄ ∩ V̄ ′) and by Lemma 4.11 we can extend it to Bi withoutmaking it touch f(M) at qi(t).

For linear maps, the Jacobian matrix is constant and equal to the coefficientmatrix. Thus, on V̄ ∪ V̄ ′ where φ is linear, we have

vi(t) = (dφ)ri(t)(ui(t)) = Dφ(ri(t)) · ui(t) = φ(ui(t)) .

Therefore, for ri(t) ∈ V̄ ∪ V̄ ′, we have

φ(ri(t) + αui(t)) = qi(t) + αvi(t), ri(t) ∈ V̄ ∪ V̄ ′, |α| ≤ ρ .

We can choose closed intervals Ii(t) ⊂ [−ρ, ρ] such that

{ri(t) + αui(t) | t ∈ [0, 1], i = 1, 2, α ∈ Ii(t)} = τ̄ ′ ,

and hence extend φ to the closed neighbourhood τ̄ ′ of A.Now, we can extend φ continuously over τ as ψ′, which is a smooth map by a

smooth approximation. We find the embedding ψ as a δ-approximation of ψ′ byTheorem 3.22, since 2n ≥ 5. The embedding ψ satisfies (II), and since 2 + n < 2nit also satisfies (III). �

This last condition is important as a property of the 2-cell σ2 = ψ(τ), which wewill need in Theorem 4.14. We will also need a certain set of sections on the tangentbundle TR2n restricted to σ2. This is taken care of in Lemma 4.13.

Lemma 4.13. Assume q and q′ are self-intersections of f of opposite types. Thenthere exist sections w1, . . . , w2n in T (R2n)|σ2 , the restriction of the tangent bundleT (R2n) of R2n to σ2, satisfying the following:

(0) For each q∗ ∈ σ2, w1(q∗), . . . , w2n(q∗) are linearly independent.

38

x1

x2

•r

•r′

Figure 4.6. Modifying the neighbourhoods V̄ and V̄ ′.

(i) For q∗ = ψ(r∗), r∗ ∈ A,

w1(q∗) = (dψ)r∗(e1), w2(q∗) = (dψ)r∗(e2) .

(ii) For q∗ ∈ B1,

w3(q∗), . . . , wn+1(q∗) ∈ Tq∗f(M1) .

(iii) For q∗ ∈ B2,

wn+2(q∗), . . . , w2n(q∗) ∈ Tq∗f(M2) .

Proof. Since ψ is an embedding, w1(q∗) and w2(q∗) are linearly independent. Set

V n−11 = {{e3, . . . , en+1}}, Vn−12 = {{en+2, . . . e2n}} ,

where {{. . . }} denotes the subspace of R2n spanned by {. . . }. For each point q∗ ofB1 put

V n−11 (q∗) = {v ∈ Tq∗f(M1) | v⊥Tq∗B1} .

Setting B1 =⋃q∗∈B1 V

n−11 (q∗), we obtain an (n − 1)-dimensional vector bundle

B1 = {B1, π1, B1} over B1. As the base space B1 is contractible, the bundle B1 istrivial. Hence, there exist linearly independent vector fields, say, w1, w3, . . . , wn+1,in B1, which define an orientation of f(M1) at each point q∗ of B1.

Similarly for q∗ ∈ B2 setting

V n−12 (q∗) = {v ∈ Tq∗f(M2) | v⊥Tq∗B2} ,

we obtain an (n − 1)-dimensional vector bundle B2 = {B2, π2, B2}, where B2 =⋃q∗∈B2 V

n−12 (q∗). Again there exist linearly independent vector fields w′2, wn+2, . . . ,

w2n in B2, which define an orientation of f(M2) at each point q∗ of B2. Here weassume w′2(q∗) to be an element of Tq∗B2, which points in the positive directionalong B2.

Let B = {σ2 × R2n, p1, σ2} be a 2n-dimensional trivial bundle over σ2. The re-striction B|B2 of B over B2 has n+1 linearly independent sections w1, w2, wn+2, . . . ,w2n. Furthermore, if q∗ = q or q∗ = q′, the w1(q∗), . . . , w2n(q∗) are linearly indepen-dent. By Lemma 4.11 we can extend the w3, . . . , wn over B2 and for each q∗ ∈ B1

39

the 2n− 1 vectors

w1(q∗), w2(q∗), w3(q∗), . . . , wn(q∗), wn+2(q∗), . . . , w2n(q∗)

are linearly independent.Recall that the self-intersections q and q′ have opposite types. By suitable choices

of w1, w3, . . . , wn+1 in B1 and w′2, wn+2, . . . , w2n in B2 we may assume that forq∗ = q or q∗ = q′, the vectors

w1(q∗), w3(q∗), . . . , wn+1(q∗), w′2(q∗), wn+2(q∗), . . . , w2n(q∗)

define the orientation opposite from the preassigned orientation in R2n.Now we can deform w2(q) and w2(q′) to w′2(q) and w′2(q′) keeping them in T (σ2)

and out of T (B1); thus, these vectors remain linearly independent of the abovevectors. Hence, the vectors

w1(q∗), w2(q∗), . . . , w2n(q∗)

define the positive orientation of R2n at q∗ = q or q∗ = q′. Therefore, we canextend the section wn+1 over B2 while keeping its linear independence with theother vectors.

Using Lemma 4.11 again we extend the sections w3, . . . , wn+1 over σ2 so thatw1, . . . , wn+1 are linearly independent at each point of σ2.

Finally, we extend the sections wn+2, . . . , w2n defined over B2 to σ2 while keepingtheir linear independence. This is possible since we may think of B2 as a smoothdeformation retract of σ2. Hence, we have shown the lemma is valid. �

As a last step before proving Whitney’s second embedding theorem, we now showhow to “add” and “remove” pairs of regular self-intersections. Thus we prove theexistence of completely regular immersions with one pair more or less than a givencompletely regular immersion f , of Figure 4.4.

The following theorem states that it is, under certain circumstances, possibleto both add and remove pairs of regular singlurarities. For the second embeddingtheorem, we only need to be able to remove pairs of regular singularities. Therefore,only a scetch is made of how to add them.

Theorem 4.14. Let f : M → R2n be an immersion, where M is an n-dimensionalmanifold and n ≥ 3. Then there exists a regular homotopy {ft|t ∈ [0, 1]} of fsatisfying the following:

(1) f0 = f .

(2) f1 is a completely regular immersion.

(3) The number of self-intersections of f1 is two more than that of f .If (i) M is orientable, n is even, and the number of self-intersections of f is

greater than |If |, or (ii) M is non-orientable or n is odd, and f has at least oneself-intersection, then there exists a regular homotopy {ft} of f such that the numberof self-intersections of f1 is two less than that of f = f0.

40

x1

x2

D1•p

•q

x1

x2

•p

•q

x1

x2

•p

•

Figure 4.7. Introducing a pair of regular self-intersections to a 1-dimensional manifold by pulling it through itself. An intermediatestep is shown, where the point p has just been pulled away fromthe x1-axis.

Proof. Notice first that by Theorem 4.5 we may assume that f is a completelyregular immersion. Take a sufficiently small n-disk Dn in f(M), Dn ⊂ f(M) ⊂ R2n,and choose distinct points p and q in Dn.

(a) We try to grasp a whole picture with n = 1. We take an embedding of D1in R2 as shown in Figure 4.7 and make two self-intersections by moving the diskaround by a regular homotopy, say, q = 0, p ∈ D1 ⊂ R1 ⊂ R2.

(b) Assume Dn is embedded in Rn ⊂ R2n. Let q = 0 and p be points of Dn withp 6= q. Assume p sits on the x1-axis. Pull up p to position some neighbourhood of xparallel to the (x1, xn+2, . . . , x2n)-plane. Next we move this neighbourhood in the(x1, xn+2, . . . , x2n)-plane through the origin 0 (at this point the disk crosses itself)making sure not to create any other intersections. The only intersections are onthe x1-axis; we have created two self intersections. We show how to eliminate twoself-intersections in case (ii) where either M is non-orientable or n is odd, and howto eliminate two self-intersections of distinct types in case (i) where M is orientableand n is even, in each case through a regular homotopy.

Take a two-cell σ2 with boundary B such that f(M) ∩ σ2 = B (Lemma 4.12).Next, deform f through σ2 in a neighbourhood of C2 in M so that the new image

of C2 will not meet B1. In this way we will remove the two self-intersections.Now, for the case of removing pair of regular singularities. Consider τ ⊂ R2 ⊂

R2n. For each point r = (a1, a2, . . . , a2n) of R2n set r∗ = (a1, a2, 0, . . . , 0) and

ψ(r) = ψ(r∗ +

2n∑i=3

aiei

)= ψ(r∗) +

2n∑i=3

aiwi(ψ(r∗)) .

41

x1

x2

A1

A2

•r

•

r′

x2 = µ(x1)

Figure 4.8. Undoing intersections. The purpose of the Whitney trick.

For each point q∗ of σ2, the vectors w1(q∗), . . . , w2n(q∗) are linearly independentand they are smooth in q∗. Therefore, ψ when considered as a map from a neigh-bourhood of σ2 to R2n has the nonzero Jacobian matrix at each point in its domain.Hence, we have the inverse ψ−1. Setting

N1 = ψ−1(f(M1)), N2 = ψ−1(f(M2)) ,

we see that N1 and N2 are contained in a neighbourhood U of τ in R2n. If wecan deform N2 in U so N2 does not intersect N1 (i.e., there exists an isotopy{it : t ∈ [0, 1]} of the inclusion map i : N2 → U such that i0 = i, i1(N2) ∩N1 = ∅),the family {ψ ◦ it} defines a deformation of f and the number of self-intersectionsof f decreases by two. Hence, in this case the proof will be complete. Set

π(x1, . . . , x2n) = (x1, 0, x3, . . . , x2n) .

By the conditions (i), (ii), and (iii) of Lemma 4.11 and the definition of ψ(r) as givenabove for r∗ ∈ A1, Tr∗N1 is in the (x1, x3, . . . , xn+1)-plane. Hence Tr∗π(N1) is alsoin the (x1, x3, . . . , xn+1)-plane. Similarly Tr∗π(N2) is in the (x1, xn+2, . . . , x2n)-plane. Hence, π(N1) ∩ π(N2) is on the x1-axis.

Let µ(x1) be a smooth function whose graph x2 = µ(x1) is the union of the set A2and the x1-axis minus the set A1 smoothed out at the points r and r′ (Figure 4.8).Take ε > 0 such that the interior of N2 consists of points whose distances fromthe (x1, x2)-plane are each less than ε. Consider a smooth function ν : R1 → R asfollows:

|ν(λ)| ≤ 1, ν(0) = 1,ν(λ) = 0 if |λ| ≥ ε2 .

Now define a map θt : R2n → R2n by

θt(r) = r − tν(x23 + · · ·+ x22n)µ(x1)e2, r = (x1, . . . , x2n) ∈ R2n .

By the definition of ν, θt is the identity map outside N2. Evidently the {θt : R2n →R2n} is a regular homotopy with θ0 = 1. When t = 1, θ1 maps the portion of N2on A2 to the halfplane x1 < 0. But π(θt(N2)) = π(N2) and π(N1) ∩ π(N2) is onthe x1-axis, and so N1 ∩ θ1(N2) must be on the x1-axis. However, θ1(N2) does notintersect the x1-axis, and so θ1(N2) must be empty.

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Furthermore, since ψ(τ)∩f(M) = B no new self-intersection will arise if we takeε sufficiently small. This concludes the proof for the case that M is orientable andn is even.

In other cases we can also remove pairs of self-intersections by regular homotopiesin just about the same way as above. To do this we only need to show that C1 andC2 can be chosen so that q and q′ have opposite types.

The case M is not orientable. If we can take C1 and C2 as above and q and q′are of opposite types we follow the previous argument. If q and q′ is of the sametype, we choose a curve C ′2 from p2 to p′2 so that C2 ∪ C ′2 reverses the orientationin M . Then q and q′ are of opposite type with respect to C1 and C ′2.

The case n is odd and M is orientable. Assume q and q′ have the same type forC1 and C2, Then choose a curve C ′1 from p1 to p′2 and a curve C ′2 from p2 to p′1 asfollows. The curve C ′i agrees with Ci near the starting point and with Cj , i 6= j,near the endpoint. The neighbourhood M ′i of C ′i is chosen in such a way that Miagrees with M ′i near the point pi and with M ′j near p′i, j 6= i. Orient Mi and M ′iwith the preassigned orientations near pi and p′i. Then q and q′ have the opposingorientations with respect to (M ′1,M ′2). �

Immersions and embeddings of non-compact manifolds are proper, in additionto what the original Definition 2.11 specifies. Continuous maps are proper if theinverse image of every compact set is compact. With this, we finally have everythingwe need for the proof of Whitney’s second embedding theorem.

Theorem 4.15 (Whitney’s second embedding theorem). Every n-dimensional dif-ferentiable manifold can be embedded in R2n.

Proof. Let M be an n-dimensional manifold. The theorem is routine for n = 1 asM1 is a finite union of S1.

Case n = 2. We can embed the sphere S2, the projective space RP 2 and theKlein bottle K2 in R4. By the classification theorem for closed surfaces, M is aconnected sum of a finite number of copies of the above surfaces. Hence, we canembed M in R4 (cf. any elementary text covering the classification of surfaces,e.g. [20]).

Let n ≥ 3. By Theorem 4.9 there exists a completely regular immersion f0 :M → R2n with If0 = 0. By Theorem 4.14 we can remove all self-intersections.Thus, we have obtained an embedding f1 : M → R2n as an injective immersion ona compact set.

Now assume that M is not compact and let (pi, qi) denote the double points off0. For each pi, take a curve γi from pi to infinity. Now define neighbourhoodsUi, for each curve γi, such that Ui intersects no other Uj or contains any other pj .The points of the Ui’s may be written as (p, r, s) with p ∈ Sn−20 , 0 ≤ r ≤ 1 and0 ≤ s < 1, such that r = 0 and 2δ ≤ s < 1 represents the curve while r = 1 or s = 0gives the boundary of Ui. Next, define the deformation

φt(p, r, s) =(p, r, [1− (1− δ)(1− r)t]s

),

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which leaves the boundary of Ui fixed. Replacing φt by a smooth deformation foreach i gives a mapping f with f(M) ⊂ f0(M) \ ∪iγi. In this way we remove allself-intersections, and f is still proper; Hence, an embedding. �

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5. Acknowledgements

For helping me with subtle but important mathematical details, the art of writinga beautiful essay, and for giving me moral support, I wish to thank my supervisorPer Åhag, he who mysteriously combines godlike mathematical and socoeu ail skills.I also wish to thank Robin Ekman, Klara Leffler and my examinator Lisa Hed, forreading my essay and giving me invaluable constructive feedback.

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References[1] Adachi M., Embeddings and immersions. Translated from the 1984 Japanese original by Kiki

Hudson. Translations of Mathematical Monographs, 124. American Mathematical Society,Providence, RI, 1993.

[2] Dieudonné J., A history of algebraic and differential topology 1900-1960. Modern BirkhäuserClassics. Birkhäuser Boston, Inc., Boston, MA, 2009.

[3] Donaldson S. K., An application of gauge theory to four-dimensional topology. J. DifferentialGeom. 18 (1983) 279-315.

[4] Donaldson S. K., One hundred years of manifold topology. History of topology, 435-447,North-Holland, Amsterdam, 1999.

[5] Montgomery D., Zippin L., Topological transformation groups. Reprint of the 1955 original.Robert E. Krieger Publishing Co., Huntington, N.Y., 1974.

[6] Freedman M. H., The topology of four-dimensional manifolds. J. Differential Geom. 17 (1982),357-453.

[7] Katz V. J., Differential forms. History of topology, 111-122, North-Holland, Amsterdam,1999.

[8] Kervaire M. A., A manifold which does not admit any differentiable structure. Comment.Math. Helv. 34 (1960) 257-270.

[9] Lee J. M., Manifolds and differential geometry. Graduate Studies in Mathematics, 107. Amer-ican Mathematical Society, Providence, RI, 2009.

[10] Lützen J., Interactions between mechanics and differential geometry in the 19th century.Arch. Hist. Exact Sci. 49 (1995), 1-72.

[11] Milnor J. W., Weaver D. W., Topology, from the differentiable viewpoint. The UniversityPress of Virginia, Charlottesville. Princeton, New Jersey, 1965.

[12] Prasolov V. V., Elements of combinatorial and differential topology. Translated from the2004 Russian original by Olga Sipacheva. Graduate Studies in Mathematics, 74. AmericanMathematical Society, Providence, RI, 2006.

[13] Prasolov V. V., Elements of homology theory. Translated from the 2005 Russian originalby Olga Sipacheva. Graduate Studies in Mathematics, 81. American Mathematical Society,Providence, RI, 2007.

[14] Scholz E., The concept of manifold, 1850–1950. History of topology, 25-64, North-Holland,Amsterdam, 1999.

[15] Simmons G. F. Introduction to topology and modern analysis. Reprint of the 1963 original.Robert E. Krieger Publishing Co., Inc., Melbourne, Fla., 1983.

[16] Steenrod N., The topology of fibre bundles. Reprint of the 1957 edition. Princeton Landmarksin Mathematics. Princeton Paperbacks. Princeton University Press, Princeton, NJ, 1999.

[17] Tu L. W., An introduction to manifolds. Second edition. Universitext. Springer, New York,2011.

[18] Whitney H., Differentiable manifolds. Ann. of Math. (2) 37 (1936), no. 3, 645-680.[19] Whitney H., The self-intersections of a smooth n-manifold in 2n-space. Ann. of Math. (2)

45, (1944). 220-246.[20] Zeeman E. C., An introduction to topology: the classification theorem for surfaces. Lecture

notes 1966. Available at: http://www.maths.ed.ac.uk/ aar/surgery/notes.htm.[21] Zisman M., Fibre bundles, fibre maps. History of topology, 605-629, North-Holland, Amster-

dam, 1999.E-mail address: miltonpersson@gmail.com

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