sign in sign up
<<
Home , Differentiable manifold, Embedding, Hassler Whitney, Immersion (mathematics), Submanifold

The Whitney theorem Milton Persson

VT 2014 Examensarbete, 15hp Kandidatexamen i matematik, 180hp Institutionen f¨ormatematik och matematisk statistik

Abstract A fundamental theorem in differential is proven in this essay. It is the embedding theorem due to , which shows that the ever so general and useful topological spaces called , can all be regarded as subspaces of some Euclidean . The version of the proof given in this essay is very similar to the original from 1944. Modern definitions are used, however, and many illustrations have been made, wherever it helps the understanding.

Sammanfattning En fundamental sats i differentialgeometri bevisas i denna uppsats. Det ¨arinb¨addni- ngssatsen av Hassler Witney, som visar att de ofantligt generella och anv¨andbara topologiska rummen kallade m˚angfalder,kan ses som delrum av n˚agotEuklidiskt rum. Versionen av beviset given i denna uppsats ¨arv¨aldigtlik originalet fr˚an1944. Moderna definitioner anv¨andsdock, och m˚angaillustrationer har gjorts, varhelst det hj¨alper f¨orst˚aelsen. Contents List of Figures 1 1. Introduction 3 2. Preliminaries 5 2.1. Manifolds 5 2.2. Immersions and 7 2.3. Bundles and polyhedra 8 3. Whitney’s first embedding theorem 13 4. Whitney’s second embedding theorem 27 5. Acknowledgements 45 References 47 List of Figures

2.1 Example of a that is not a differentiable 6 2.2 A mapping on manifolds and the corresponding mapping on Euclidean spaces. 7 2.3 Example of what is not the of an . 8 2.4 A 2- with an illustration of how it is generated by its points. 11 3 2.5 2-dimensional polyhedron in R consisting of seven faces. 12 3.1 Dependency diagram for the to the proof of Whitney’s first embedding theorem. 13 3.2 Illustration for proof of Lemma 3.11 18 3.3 Illustration for proof of Lemma 3.12. 19 4.1 Dependency diagram for the path to the proof of Whitney’s second embedding theorem. 27 4.2 Mappings of Theorem 4.5. 30 4.3 A completely regular immersion with one self-intersection. 32 4.4 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. 36 4.5 Defining vector fields necessary for the Whitney trick. 38 4.6 Modification of neighbourhoods, part of the Whitney trick. 39 4.7 Introducing a pair of regular self-intersections to a 1-dimensional manifold. 41 4.8 Undoing intersections. The purpose of the Whitney trick. 42

1

1. Introduction As every one knows an n-dimensional differentiable manifold is defined to be a M together with a differentiable structure D such that the induced is Hausdorff, second countable and such that M is locally Euclidean of n. Some, for example [9], assume only paracompactness instead of the stronger assumption of second countability. We can not proceed in this manner since we want to apply Sard’s theorem. It should also be emphasized that not every set admits a differentiable structure (see e.g. [8] or [3, 6]). This crucial point is something that most textbooks in differentiabl geometry forget to mention. For convenience we shall assume that our manifolds are C∞-smooth, even if the proofs go through for C1-smooth manifolds. In [14], Scholz argue that the concept of manifolds can be traced back to Lagrange and his “M´ecanique analitique” published in 1788, other interpret it differently. We shall not here entangle in this fight. Instead, we refer the interested reader to [2, 4, 7, 10, 21] and the references therein. On the other hand, most agree on that it was who in “Idee der Riemannfl¨ache” published in 1913 introduced for the first time what we today know as real manifolds. The modern definition of manifolds is completely intrinsic, so we can not a priori N view a manifold globally as a subset of R for some N. In other words, we can not N assume that the manifold can be embedded in R . This is one of the most crucial points of our definition. The main of this essay is the American Hassler Whitney who in 1936 proved that any n-dimensional differentiable manifold 2n+1 can be embedded in R ([18]). The proof of this theorem is included in most introductory textbooks on differential geometry (see e.g. [9, 17]), and it has been considerably simplified over time. But the original work of Whitney is still to this day 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 of this essay. He prove the following:

Theorem 4.15. Every n-dimensional differentiable manifold can be embedded in 2n R . This theorem is sharp as it is shown by the example that the 2-dimensional pro- 4 3 jective RP 2 can be embedded in R , but not in R . It is hard to grasp the importance of Whitney’s embedding theorem and its proof. One thing that might indicate this is that Medal laureates , , Grigori Perelman, and have all, directly or indirectly, been inspired by Theorem 4.15 and its proof. Another thing is that it yields the foundation of a whole field of mathematics known as . Finally, to quote the French legend Jean Alexandre Eug`eneDieudonn´ein 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 in the literature. Except for the original articles by Whitney it can be found in [1], a Japanese translation of Whitney’s original work that in 1993 was translated back

3 2n+1 into English. Another rare exposition was written by Prasolov; the case R can be found in [12], and the generalization can be found in [13]. In this essay we follow [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 is what 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, and it is only valid for dimension 5 or higher. Therefore, in the proof of Theorem 4.15 for the lower we shall need to use the classification theorems for lower dimensional manifolds. The overview of this thesis is as follows. In §2, we give some necessary background to 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 theorem 2n+1 R , 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, and its trick, it should be pointed out that some depth of detail is ignored. The main thing omitted is in Lemma 4.11 when we refer to [16] for the details of extending a over |K0| to |K| in the (m − i)-dimensional over |K| (see §2.3 for definition).

4 2. Preliminaries Important or reoccurring notions and concepts of differential geometry are pre- sented in the three subsections that follow. Concepts specific to certain lemmas and theorems are defined when necessary. 2.1. Manifolds. Definition 2.1. A set M is locally Euclidean of dimension n if every point p ∈ M has a neighbourhood U such that there is a bijection φ from U into an open subset n of Rn. We call the pair (U, φ : U → R ) a chart, U a coordinate neighbourhood, and φ a coordinate map on U. n n n n Definition 2.2. Two charts (U ⊂ R , φ : U → R ), (V ⊂ R , ψ : V → R ) are C∞-compatible if the two maps φ ◦ ψ−1 : ψ(U ∩ V ) → φ(U ∩ V ) , ψ ◦ φ−1 : φ(U ∩ V ) → ψ(U ∩ V ) , n n are both C∞-smooth as maps R → R . n Definition 2.3. An on a set M is a collection S = {(Uα, φα : U → R )} that satisfy the following: (i) The coordinate neighbourhoods cover M, i.e. [ M = Uα . α n (ii) Every image of the form φα(Uα ∩ Uβ) is open in R . (iii) They are pairwise C∞-compatible. The neighbourhoods of the charts from an atlas S on a set M constitute a basis for a 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 can find the requirements necessary, on an atlas S , for the induced topology to be Hausdorff and second countable. For us, the important part is that there exist such requirements. Definition 2.4. Two atlases on a set M are said to be equivalent if their union is also an atlas on M. An equivalence of atlases on M is called a differentiable structure on M. The pair (M, D) of a set M and a differentiable structure D on M is called a differentiable manifold if the topology induced by D makes (M, D) a Hausdorff, locally Euclidean, second countable topological space. We will often write M instead of (M, D), when the differentiable structure is understood. Remark. Manifolds do not have to be connected. Manifolds are said to be compact if every component is compact. They are said to be non-compact no component is compact. Until the very last part of Whitney’s second embedding theorem we will only consider compact manifolds. Let us now give some examples of differentiable manifolds, as well as sets that cannot be made into differentiable manifolds.

5 n Example 2.5. R with differentiable structure represented by n n {(R , id)}, is a differentiable manifold, and so is every open U ⊂ R , with repre- n n sentatives {(U, id|U )}. Here, id : R → R is the identity map. Example 2.6. The cross given by 2 {(x, 0) | x ∈ (−1, 1)} ∪ {(0, y) | y ∈ (−1, 1)} ⊂ R , is not a differentiable manifold (see Figure 2.1). The reason is that there is no bijection between a neighbourhood of the intersection and an open subset of any Euclidean space (see e.g. [17]).

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

An alternative definition of differentiable manifold is as a set M together with a so called maximal atlas. Given any atlas on M, a maximal atlas is formed as the union of all equivalent atlases, which is the set of all compatible charts. This is precisely what the differentiable structure consists of. Thus, given an atlas on M, we obtain the same differentiable manifold whether we give it the corresponding maximal 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 equivalence classes 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 different differentiable structures (see [6]). Of course, there is no guarantee that there exists a differentiable structure for an arbitrary set M. Even if M is a Hausdorff, locally Euclidean, second countable topological space from the outset, its topology may not be the same as that induced by a differentiable structure. In this essay, however, we will simply assume that this problem does not occur. We shall now introduce the concept of . There are various definitions in 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 subset k n of M. Regard R , 0 ≤ k ≤ n, as a subspace of R , k n R = {(x1, . . . , xn) ∈ R | xk+1, . . . , xn = 0} .

Now assume that we can choose a representative S = {(Vj, ϕj)} of D such that for each nonempty Vj ∩ A, the restriction k n ϕj|Vj ∩A : Vj ∩ A → R ⊂ R

6 M1 f M2 · x · U U1 2

h−1 h2 1 −1 h2 ◦ f ◦ h1

h2(U2) · h1(U1) · n R

Figure 2.2. A mapping f of n-dimensional manifolds M1 and M2. The point h1(x) ∈ U1 ⊂ M1 is mapped to h2(f(x)) by the −1 n composition h2 ◦ f ◦ h1 : h1(U1) → R .

k is a into an open subset of R and the chart (Vj ∩A, ϕj|Vj ∩A) is said to be adapted to A. Thus, we have an atlas SA = {(Vj ∩ A, ϕj|Vj ∩A)} representing a differentiable structure DA. We say that (A, DA) is a submanifold of (M, D) of dimension k. 2.2. Immersions and embeddings. In this section we define immersions and embeddings and all that is necessary for those definitions. We shall also give some examples of sets that can, and that cannot, be the images of immersions or embeddings.

Definition 2.8. Let (M1, D1) and (M2, D2) be differentiable manifolds. A mapping f : M1 → M2 is smooth at a point x ∈ M1 if there exists charts (U1, h1), from some atlas 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. −1 −1 (iii) The map h2 ◦f ◦h1 : h1(U1 ∩f (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 f and f −1 are smooth.

Definition 2.10. Let M1 and M2 be n-dimensional manifolds and f : M1 → M2 a smooth map. For x in M1 choose charts (U1, h1) and (U2, h2) about x and f(x). We define the of f at x to be the rank of the Jacobian matrix of the map −1 −1 h2 ◦ f ◦ h1 : h1(U1 ∩ f (U2)) → h2(U2) at h1(x).

7 3 Figure 2.3. The double cone in R is not the image of an immersion.

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

Definition 2.11. A smooth map of manifolds f : M1 → M2 is an immersion if the rank of f at each point of M1 is the same as the dimension of M1. If f is both an immersion and a homeomorphism, then it is an embedding. If there exists an embedding of M1 into M2, then we say that M1 is embedded in M2. 2 3 Example 2.12. The map f : R → S1 × R → R defined by f(u, v) = (v cos u, v sin u, v) , is smooth but not an embedding, since it intersects itself at (u, v) = (u, 0) (see Figure 2.3). Examining f more closely, as follows, we see that it is not even an 2 3 2 immersion. Since f : R → R is a map of R , we can choose identity maps as the coordinate maps in the definition of rank (Definition 2.10). Thus, its rank is the rank of its Jacobian matrix cos u −v sin u Df(u, v) = sin u v cos u  . 1 0 which has rank 1 < 2 whenever v = 0.

2 Example 2.13. The mapping α : R → R defined by, x 1 y = x − , z = (1 + x2) (1 + x2) with graph in Figure 4.3 on p. 32, is an immersion but not an embedding since it is not injective, passing through (0, 1/2) twice. We shall get back to this example in (4.2).

2.3. Bundles and polyhedra. This section consists of definitions necessary for Lemma 4.11, which will in turn be used in Lemma 4.12 and Lemma 4.13. For a full description of how the lemmas and theorems of §4 depend upon one another, see Figure 4.1. First we shall introduce the concept of topological transformation groups. We then need the theory of topological groups. We refer to [5] for the necessary back- ground.

8 Definition 2.14. Let G be a topological , 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, g0 ∈ G and y ∈ Y , η(gg0, y) = η(g, η(g0, 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 of dimension n is the set of all non- nn singular n × n matrices GL(n, R) = {A ∈ R | det(A) 6= 0}. The orthogonal group nn of dimension n is the set of all orthogonal n×n matrices O(n) = {A ∈ R | AT A = n I} ⊂ GL(n, R). Both of these are topological transformation groups of R , their identity element is the identity matrix, e = I, and the continuous map η corresponds to 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 X is the base space. The member p : B → X is a surjective continuous map called the projection map of B. (2) The member Y is also a topological space; Y is the fiber of B. The member G 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 a homeomorphism

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

φj,x(y) = φj(x, y), y ∈ Y

gives a homeomorphism of Y ,

−1 φj,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

−1 gji(x) = φj,x ◦ φi,x .

Then gji is continuous; we say that the gji’s are coordinate transfor- mation or transition functions of B. −1 We write Yx for p (x); Yx is the fiber over x.

9 Definition 2.17. For any two maps p : E → M and p0 : E0 → M with the same target space M, a map φ : E → E0 is said to be fiber-preserving if φ(p−1(x)) ⊂ p0−1(x) for all x ∈ M. Definition 2.18. A surjective smooth map p : B → X of manifolds is said to be locally trivial of rank r if (i) each fiber p−1(x) has the structure of a of dimension r; (ii) for each x ∈ X, there are an open neighbourhood U of x and a fiber- r preserving diffeomorphism φ : p−1(U) → U × R such that for every x ∈ U the restriction −1 r φ|p−1(x) : p (x) → {x} × R is a vector space . n Definition 2.19. A bundle B = {B, p, X, Y, G} in which Y = R and G = GL(n, R) acts on Y by the usual linear transformations is called an n dimensional . If p : B → X is locally trivial of rank r then B is said to be a vector bundle of rank r. r Definition 2.20. Given a manifold M, let p1 : M × R → M be the projection to r r the first factor, i.e. p1(m, x) = m, where m ∈ M and x ∈ R . Then M × R → M is a vector bundle of rank r, called the product bundle of rank r over M. The vector −1 r space structure on the fiber p1 (x) = {(x, v) | v ∈ 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 M r to the product bundle M × R is called a trivial bundle. The remainder of this section consists of the definition of polyhedra and all that is necessary for that definition. The importance of polyhedra in this essay comes from the fact that all differentiable manifolds are homeomorphic to a polyhedron. In other words, every differentiable manifold is triangulable. Definition 2.22. Let P ,...,P be affinely independent points in m, m ≥ n, −−−→ 0 −−−→n R which means that P0P1,..., P0Pn are linearly independent. We call the set ( ) −−→ −−→ −−→ m OX = λ0OP0 + ··· + λnOPn, |P0P1 ...Pn| = X ∈ R λ0 + ··· + λn = 1, an n-simplex. The n is the dimension of the simplex (see Figure 2.4).

q n Definition 2.23. With a subset {ik | 0 ≤ ik ≤ n}k=0 of an index set {i}i=0, any set of 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 = |Pi0 Pi1 ...Piq | , called a q-face of σ. If τ is a face of σ, then we write τ ≺ σ or σ τ .

10 λ0 = 1/5, λ1 = 3/5, λ2 = 1/5

P1 P1 • •

P2 P2 • •

• • O O

• • P0 P0 1 2 3 4

λ0 = 1/5, λ1 = 3/5, λ2 = 1/5 |P0P1P2|

P1 P1 • • X P2 P2 • •

• • O O

• • P0 P0

Figure 2.4. A 2-simplex with an illustration of how it is generated by 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 if it satisfies 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 dim K. 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 all simplices belonging to K is a polyhedron of K denoted by |K|, [ n |K| = σ ⊂ R . σ∈K

11 P6 • P7 •

• P1

P2 • • P4 |P0P1P2|

• • O P5

• P0 • P3

 |P0P1P2|, |P0P1P3|, |P1P2P6|, |P1P6P7|, |P1P4P7|, |P1P4P5|, |P1P3P5|

3 Figure 2.5. 2-dimensional polyhedron in R consisting of seven faces.

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

12 3. Whitney’s first embedding theorem We have included a diagram below (Figure 3.1) which gives an overview of this section and how we will reach the proof of Theorem 3.22.

The first embedding theorem. Theorem 3.22

Lemma 3.20 Lemma 3.9

Theorem 3.10 Theorem 3.17 Lemma 3.13 Lemma 3.18 Lemma 3.21

Corollary 3.8 Lemma 3.12

Corollary 3.6 Lemma 3.11

Lemma 3.5

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

We follow the diagram in Figure 3.1 by proving the three theorems as soon as we can, 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 + b = x, ka − bk n = 2r . 2 R 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 its corollaries.

Definition 3.2. A cover of a topological space S is a collection of sets {Uα}α∈A S such 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 set in the refinement is a subset of at least one of the sets in the original cover. A cover of a topological space is said to be locally finite if every point has a neighbourhood that intersects only a finite number of sets from the cover. Definition 3.3. A topological space X is said to be paracompact if every open cover of X has a refinement which is locally finite and open.

13 n Definition 3.4. A subset A of R is said to have measure zero if for an arbitrary ε > 0 there exists an open cover of A, ∞ [ n A ⊂ C (xi, ri) , i=1 such that ∞ X n ri < ε . i=1 n n Lemma 3.5. Let f : U → R be a smooth map, with U an open subset of R . If A 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 have measure zero. Then, for an arbitrary ε > 0, there exists an open covering of A ∩ C, ∞ [ n A ∩ C ⊂ C (xi, ri) , i=1 such that ∞ X n ri < ε . i=1 Furthermore, if

 ∂f  b = max i x∈C,i,j ∂xj x then by applying the mean value theorem in several variables to each component of f, we get

p 2 2 kf(x) − f(y)k n ≤ b + ··· + b kx − yk n = R R n2 terms

= bnkx − yk n , x, y ∈ C. R With this we have found that n n f(C (xi, ri)) ⊂ C (f(xi), bnri) . Thus ∞ ! ∞ [ n [ n f(A ∩ C) ⊂ f C (xi, ri) = f(C (xi, ri)) ⊂ i=1 i=1 ∞ [ n ⊂ C (f(xi), bnri) . i=1 This means that f(A ∩ C) has measure zero, since ∞ ∞ X n n n n n n n X n n n n 2 b n ri = 2 b n ri < 2 b n ε . i=1 i=1

14 Finally, A can be covered with a countable number of n-cubes such as C, so the measure of f(A) must be zero.  n p Corollary 3.6. With n < p let U be an open subset of R , and let f : U → R be a smooth map. Then f(U) has measure zero. p−n p Proof. We may think of U × R as an open subset of R . Hence, we define p−n p g : U × R → R by g = f ◦ p1, i.e.

p−n p1 f p g : U × R −→ U −→ R , ∞ where p1 is the projection into the first component. As a composition of C maps, ∞ ∼ p−n g too, is a C map. But U = U × {0} ⊂ U × R has measure zero, and so p f(U) = g(U × {0}) has measure zero in R .  In Definition 3.4 we defined the concept of measure zero for a subset of Euclidean space, but we must also define it for manifolds. Definition 3.7. Let M be an n-dimensional manifold with associated differentiable structure D, and let A be a subset of M. We say that A has measure zero if n ϕα(Uα ∩ A) ⊂ R 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) pn pn is bijective to R , we give the usual topology of R 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 may assume that E =  A0 B0  with A ∈ GL(k, ). Then there exists an ε > 0 such 0 C0 D0 0 R that det(A) 6= 0 if A is an element of GL(n, R) such that the of each entry 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 form AB E = [ CD ], where A is a (k × k)-matrix such that the absolute value of each entry of 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 I 0  AB  AB  k = XIp−k CD XA + CXB + D is equal to the rank of E for an arbitrary ((p − k) × k)-matrix X. Now setting X = −CA−1 we see that the above matrix becomes AB  . 0 −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. m Let W be an open subset of R , where m = pn − dim(D) = pn − (p − k)(n − k) = k(p + n − k) ,

( )   AB The absolute value of each entry W = ∈ M(p, n; R) . C 0 of A − A0 is less than ε. Then the correspondence AB AB  7→ C 0 CCA−1B defines a diffeomorphism between W and the neighbourhood U ∩ M(p, n; k) of E0 in M(p, n; k). Therefore, M(p, n; k) is a k(p+n−k)-submanifold of M(p, n; R).  n p Theorem 3.10. If f is a smooth map of an open subset U of R into R , with p ≥ 2n, then for an arbitrary ε > 0, there exists a (p×n)-matrix A = [aij] satisfying the following:

(i) |aij| < ε. p (ii) The map g : U → R defined by g(x) = f(x) + Ax is an immersion. Proof. We must show that the rank of Dg(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. Consider the 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)×U is 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 be zero in M(p, n, R). Throughout all of this, no assumptions had to be made about the size of the entries of A. Thus, there is a matrix A which satisfies (i) and which is not contained in any of the Fk’s, k = 0, 1, . . . , n − 1, i.e. which does not have rank 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 be used in Lemma 3.18 and Lemma 3.21, Theorem 3.17 and Theorem 4.5. One of the properties of this atlas is that the neighbourhoods constitute a locally finite refinement of an arbitrary open covering of the underlying manifold. Thus, the underlying manifold must be paracompact. In the following, we use int A to denote the 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 proceed to construct a sequence of compact sets {Ai} such that ∞ [ X = Ai , i=1 and

Ai ⊂ int Ai+1 .

The Ai’s obtain these properties if we set A1 = U 1 and

Ai+1 = (U1 ∪ · · · ∪ Uk) ∪ U i+1 , where k is the smallest 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 they can be covered by finitely many of the Wj’s. Therefore, there exist a finite number of sets Vir such that Ss (i) Ai+1 \ int Ai ⊂ r=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 that P is a locally finite refinement of the arbitrary covering W of X. Thus, X is paracompact.  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}. n (iii) hj(Vj) = C (3). S −1 n (iv) M = j Wj, where Wj = hj (C (1)).

17 A5 A4 A3 A2

V33 V32 V31

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

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

A1,A2,..., such that ∞ [ M = Ai , i=1 and

Ai ⊂ int Ai+1 .

With the additional requirement that

−1 n −1 n Ai+1 \ int Ai ⊂ ∪jhj (C (1)) including A1 ⊂ ∪jhj (C (1)) ,

−1 n the last condition (iv) is satisfied, as follows. What we want is ∪jhj (C (1)) = M = ∪iAi and since Ai ⊂ int Ai+1 we have [ [ Ai+1 \ int Ai = Ai , i i see Figure 3.3, so

[ [ [ −1 n M = Ai = Ai+1 \ int Ai ⊂ hj (C (1)) . i i j But [ −1 n [ −1 n [ hj (C (1)) ⊂ hj (C (3)) = Vj = M j j j

18 A3 \ int A2

A2 \ int A1

A3 A2 A1

Figure 3.3. Illustration of how ∪iAi+1 \ int Ai = ∪iAi. so [ −1 n M ⊂ hj (C (1)) ⊂ M j i.e. [ −1 n [ M = hj (C (1)) = Wj . j j  If we can prove there is a mapping with some desirable properties on just a neighbourhood of some point in a manifold, then we can extend it to obtain a map with the same properties on the entire manifold using what is known as a cut-off function. We simply multiply the mapping with a function that is 1 where the mapping is defined and has compact containing that domain. In this essay the following special case of such a function will suffice. n Lemma 3.13. There exists a smooth function ϕ : R → R satisfying the following: (i) ϕ(x) = 1, x ∈ Cn(1). (ii)0 < ϕ(x) < 1, x ∈ Cn(2) \ Cn(1). n (iii) ϕ(x) = 0, x ∈ R \ Cn(2). We call ϕ a cut-off function.

Proof. Define a function λ : R → R by e−1/x, x > 0, λ(x) = 0, x ≤ 0. and set λ(2 + x)λ(2 − x) ψ(x) = . λ(2 + x)λ(2 − x) + λ(x − 1) + λ(−x − 1) Now define ϕ by n Y ϕ(x1, . . . , xn) = ψ(xi) . i=1 

19 Definition 3.14. Let f0 : X → Y and f1 : X → Y be smooth maps. A from 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 closed subset and if F (a, t) = f0(a) = f1(a) for all a ∈ A and all t ∈ [0, 1] then we say that f0 is homotopic to f1 relative to A and we write f0 ' f1 (rel A). We sometimes write 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 all t ∈ [0, 1] then we say that it is a and that f0 is regularly homotopic to f1 and write f0 ' f1. To be regularly homotopic relative to some closed subset r A ⊂ X is defined in the obvious way. Remark. In some texts, regular is the same as full rank. Then immersions are regular 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 space with metric d, and let δ : X → R+ be a . We say that g : X → Y is a δ-approximation of f : X → Y if d(f(x), g(x)) < δ(x) for all x in X.

p Now, given a smooth map f : M → R (p ≥ 2n, n as the dimension of M), we prove the existence of an immersion as a δ-approximation of f. Later we will prove p the existence of an embedding, given an immersion M → R (p ≥ 2n + 1). Thus, p given a smooth map M → R (p ≥ 2n + 1) we prove the existence of an embedding p M → R (p ≥ 2n + 1), i.e. Whitney’s first embedding theorem, which in turn is used in the proof of his second. p Theorem 3.17. Let M be an n-dimensional manifold, and let f : M → R be p a smooth map, with p ≥ 2n. There exists an immersion g : M → R which is a δ-approximation of f. In , 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 open neighbourhood U of N. The family {U, M \N} is an open cover of M. We select an atlas D0 = {(Vj, hj)} according to Lemma 3.12, which is a locally finite refinement n −1 n of {U, M \ N} with hi(Vi) = C (3) and Wi = hi (C (1)). We next set Ui = −1 n hi (C (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 set

εi = min δ(x) . x∈U i

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 S j≤0 W j since W j ⊂ Vj ⊂ U for j ≤ 0. p Induction hypothesis. Assume next that fk−1 : M → R is a smooth map having S rank n on Nk−1 = j

which assigns DFA(x) to (x, A) is continuous and the subset M(p, n; n) consisting of matrices of rank n is open in M(p, n; R). But since −1 Φ(K, 0pn) = {D(fk−i ◦ hk (x))|x ∈ K} ⊂ M(p, n; n) we have Φ(K,A) ⊂ M(p, n; n) for a small enough A, by which we mean that the largest entry of A should be small enough. (ii) The desired A should also be sufficiently small that

εk n kAxk n < , x ∈ C (3) . R 2k −1 (iii) Finally, by Theorem 3.10 we may take A so small that fk−1 ◦ hk (x) + Ax has rank n on Cn(2). p Having chosen the desired A as above, we define for each k a map fk : M → R by  fk−1(x) + ϕ(hk(x))Ahk(x) if x ∈ Vk, fk(x) = (3.1) fk−1(x) if x ∈ M \ U k .

The map fk is well defined, i.e. for a point x ∈ Vk ∩ (M \ U k) = Vk \ U k, we have

fk−1(x) + ϕ(hk(x))Ahk(x) = fk−1(x) .

The rank of fk is by (i) n on Nk−1 and by (iii) n on Uk. Therefore, the rank of fk S k is n on Nk = j

21 p Define g : M → R by g(x) = limk→∞ fk(x). This means the following. Recall that [ M = Wi , i and [ N0 ⊂ N1 ⊂ N2 ⊂ ...,Nk−1 = W j , j

For an arbitrary point x of M, since the {Vi} is locally finite, there exists a neigh- bourhood U(x) of x such that U(x) ∩ Vj 6= ∅ only for a finite number of j’s, the largest among which we call k; then

g(x) = fk(x) = fk+1(x) = ..., by the definition of the fk, (3.1). S Clearly, the map g is smooth and of rank n on k Nk = M. Moreover, g is a δ-approximation of f. The construction of fk implies immediately that fk is homotopic to fk−1 while N is kept fixed. Hence, f is homotopic to g with N fixed.  Proving the existence of an embedding, given an immersion, has been divided into the three lemmas Lemma 3.18, Lemma 3.20 and Lemma 3.21, that remain before Whitney’s first embedding theorem, Theorem 3.22. First we prove the existence of p an injective immersion as an approximation of a given immersion M → R (p ≥ 2n+ 1), which is Lemma 3.18. Then, Lemma 3.20 tells us what requirements an injective immersion has to fulfill in order to be an embedding. Lastly, in Lemma 3.21, we show that our injective immersion actually fulfills these requirements.

p Lemma 3.18. Let f : M → R be an immersion of an n-dimensional manifold M, with 2n < p. There exists an injective immersion g which is a δ-approximation of f. Furthermore, if f is injective in an open neighbourhood U of a closed subset N of M, we can choose g such that g is regularly homotopic to f relative to N. Proof. On one hand, because f is an immersion, there exists an open covering

{Uα} of M such that f|Uα is an embedding for each α. On the other hand, the set {U, M \N} is another covering of M. Together these coverings form a third covering with respect to which we consider an atlas {(Vj, hj)} as given in Lemma 3.12. n Furthermore, using the cut-off function ϕ : R → R of Lemma 3.13, we define ϕi : M → R by  ϕ ◦ hj(y) if y ∈ Vj , ϕj(y) = (3.2) 0 if y∈ / Vj .

22 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. p Induction hypothesis. Assuming next that we have an immersion fk−1 : M → R S p which is injective on j

fk(x) = fk−1(x) + ϕk(x)bk , (3.3) p where bk is a point of R , which we select as follows: (i) bk is small enough to make fk an immersion. k (ii) bk is small enough for fk to be a δ/2 -approximation of fk−1. (iii) Let N be a 2n-dimensional subset of M × M,

N = {(x, y) ∈ M × M | ϕk(x) 6= ϕk(y)} . Evidently N is open in M × M. p Induction step. Now consider the following smooth map Φ : N → R , −(f (x) − f (y)) Φ(x, y) = k−1 k−1 . ϕk(x) − ϕk(y) p By assumption 2n < p, and so the measure of Φ(N) in R is zero. Hence, we want bk to be outside Φ(N). If bk is small enough to satisfy the conditions (i) and (ii) it will work for the condition (iii). It is by Theorem 3.10 that condition (i) can be satisfied. The map fk satisfies  ϕk(x) − ϕk(y) = 0 , fk(x) − fk(y) = 0 if and only if (3.4) fk−1(x) − fk−1(y) = 0 .

The ‘if part’ follows directly from the definition (3.3) of the fk,

fk(x) − fk(y) = (fk−1(x) − fk−1(y)) + (ϕk(x) − ϕk(y))bk = 0 . For the converse, assume that

0 = fk(x) − fk(y) = (fk−1(x) − fk−1(y)) + (ϕk(x) − ϕk(y))bk .

Then if ϕk(x) − ϕk(y) 6= 0 we have

−(fk−1(x) − fk−1(y)) bk = ∈ Φ(N) , ϕk(x) − ϕk(y) a contradiction. Hence, we must have ϕk(x) − ϕk(y) = 0. Therefore, we also get fk−1(x) − fk−1(y) = 0. p Define g : M → R by

g(x) = lim fk(x) . k→∞

23 This means the following. We have  ϕ ◦ hk(x) if x ∈ Vk , ϕk(x) = c 0 if x ∈ Vk , and, therefore,  fk−1(x) + ϕ ◦ hk(x), if x ∈ Vk , fk(x) = c fk−1(x), if x ∈ Vk . 0 Since {Vj} is locally finite, any given x ∈ M lies only in a finite number of Vks. For c all other k, x ∈ Vk 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 that g(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 = ∅. It remains 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 k > 0.

From the former equation we get f(x) = f(x0). Therefore, x and x0 do not belong to the same Vj. But from the latter equation we see that if x ∈ Vk for k > 0, then x0 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 regularly homotopic relative to N, and hence the same is true for f and g.  p Definition 3.19. Let M be a manifold and let f : M → R be a continuous map. We say that the set    y = limn→∞ f(xn) for some  p ∞ L(f) = y ∈ R sequence {xn}n=1 ⊂ M which  has no limit point in M.  is the limit set of f. n Lemma 3.20. If f : M → R is an injective immersion of the n-dimensional manifold M then n (i) f(M) is a closed subset of R 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) is closed then limn→∞ f(xn) ∈ f(M). Now, assume that L(f) ⊂ f(M) and assume that y ∈ f(M) and define a sequence ∞ n {xn}n=1 in M such that f(xn) ∈ C (y, 1/n). Then

f( lim xn) = lim f(xn) = y . n→∞ n→∞

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 that L(f) ∩ f(M) 6= ∅ and that y ∈ L(f) ∩ f(M). Then

y = lim f(xn) , n→∞ because y ∈ L(f). Furthermore, since f is continuous, we have

f( lim xn) = lim f(xn) . n→∞ n→∞ Therefore

f( lim xn) = y n→∞ 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 contradicts f(M) ∩ L(f) = ∅.  n Remark. Note that for injective immersions f : M → R to be embeddings with n closed images f(M) ⊂ R , we need both L(f) ⊂ f(M) and L(f)∩f(M) = ∅, which is only possible if L(f) = ∅. Hence, Lemma 3.21. Lemma 3.21. Let M be an n-dimensional manifold. Then there exists a smooth p map f : M → R with L(f) = ∅.

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

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 has no limit point. Then for any m > 0 there exists an integer n > 0 with xn ∈/ W 1 ∪ · · · ∪ W m, and so xn ∈ W j for some j > m. Therefore, f(xn) > m and ∞ the sequence {f(xn)}n=1 has no limit point.  Observe that we have only showed the existence of a smooth map with empty limit set, and not an injective immersion with empty limit set. It is for this reason alone that the immersion of Theorem 3.17 and the injective immersion of Lemma 3.18 are approximations of the given smooth map and immersion, respec- tively. That way, the injective immersion is an approximation of the smooth map 2n+1 and so has empty limit set too. To see this, take a smooth function g : M → R 2n+1 with L(g) = ∅ and an injective immersion f : M → R that is a δ-approximation

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

kg(x ) − yk 2n+1 > ε k R for all N ≥ 1. Then, we can choose the continuous map δ : M → R+ as a constant map 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- 2n+1 entiable manifold can be embedded in R as a closed subset. 2n+1 Proof. Let M be an n-dimensional manifold. Let f : M → R be a smooth map with L(f) = ∅. Let δ : M → R+ be the constant map δ(x) = ε > 0. Then by 2n+1 Theorem 3.17 there exists an immersion g : M → R which is a δ-approximation of f. Furthermore, by Lemma 3.18 there exists an injective immersion h : M → 2n+1 R 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 an 2n+1 embedding, and so h(M) is a closed subset of R . 

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

Lemma 3.12 Theorem 4.9

The second Theorem 4.5 Theorem 4.15: embedding theorem.

Theorem 3.17 Theorem 4.14

Lemma 4.12 Lemma 4.13

Theorem 3.22 Lemma 4.11

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

We begin by defining , and .

Definition 4.1. For an n-dimensional manifold M, a tangent vector vp at p ∈ M ∞ ∞ is 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. The tangent bundle of M is the disjoint union of all tangent spaces: [ TM = TpM. p∈M 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 f at each point x in f −1(y) is p; otherwise, y is a critical value.

27 2n Definition 4.3. Let M be an n-dimensional manifold and let f : M → R be an immersion. 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,   2n (df)p1 Tp1 M ⊕ (df)p2 Tp2 M = TqR , (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 of f, 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 Cartesian product. Thus,

(df)p1 (Tp1 M) ⊕ (df)p2 (Tp2 M) =

={(Df(p1) · v1, Df(p2) · v2) | v1 ∈ Tp1 M, v2 ∈ Tp2 M} = ( ) h i T = Df(p1) Df(p2) (v , v ) (v , v ) ∈ T M × T M . 1 2 1 2 p1 p2

2n 2n 2n Since TqR = R for all q ∈ R , equation (4.1) becomes ( ) h i 2n Df(p1) Df(p2) v v ∈ T M × T M = . p1 p2 R

Remark. With these rewritings we see that the dimension of the image must be twice that of the domain for there to exist regular self-intersections: For the column 2n vectors of the above matrix to span R there must be 2n of them (n for each Jacobian matrix) and they must have 2n components. The only other way for a mapping to be completely regular is for it to have no self-intersections. We prove the second embedding theorem by first proving the existence of a 2n completely regular immersion of an n-dimensional manifold M to R , Theorem 4.5. In Theorem 4.9, we show how to modify a completely regular immersion so as to introduce new regular self-intersections. With this we can ensure that there are just as many regular self-intersections of positive type as of negative type (see Definition 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 an 2n embedding M → R when M is compact. To prove the existance of a completely regular immersion we shall use Sard’s theorem which we state here as Theorem 4.4. A proof can be found in [11]. p n Theorem 4.4. Let f : U → R be a smooth map, defined on an U ⊂ R , and let

C = {x ∈ U | rank dfx < p} . p Then the image f(C) has measure zero in R . 2n Theorem 4.5. Let M be an n-dimensional manifold, and let f : M → R be 2n a smooth map. There exists a completely regular immersion g : M → R which is a δ-approximation of f. Furthermore, if f is an immersion that is compeletely regular on some open neighbourhood U of a compact set N, then g may be chosen to be regularly homotopic to f relative to N. Proof. By Theorem 3.17 there exists an immersion f¯ which is a δ/2-approximation ¯ ¯ of f with f|N = f|N . The map f is completely regular on some open neighbourhood 2n 0 2n of N. Now, select an atlas of R , S = {(C (xi, 1), ψi)|i = 1, 2,... }, where 2n 2n ψi : C (xi, 1) → C (1). The family ¯−1 2n U = {U, (M \ N) ∩ f (C (xi, 1)) | i = 1, 2,... } is an open covering of M. We take an atlas {(Vj, hj)} of M given by Lemma 3.12 with respect to U , such that for each Vj the following conditions hold. ¯ 2n (i) f|Vj : Vj → R is an embedding. 2n (ii) For some λj such that f(Vj) ⊂ C (xλj , 1), there exists a diffeomorphism 2n 2n ϕj : C (1) → R with ¯ 2n n ϕj ◦ ψλj ◦ f(Vj) ⊂ C (1) ∩ R . n n 2n 2n (We think of R as R = {(x1, . . . , xn, 0,..., 0) ∈ R } ⊂ R .) (See Figure 4.2.)

Re-indexing the {(Vi, hi)}’s, as we did in the proof of Theorem 3.17, we assume

i ≤ 0 if and only if Vi ⊂ U,

i > 0 if and only if Vi ⊂ M \ N. 0 We write ϕj = ϕj ◦ ψλj . We shall construct g inductively. ¯ Induction basis. Set g0 = f. 2n Induction hypothesis. Assume that we have defined gj : M → R with gj|N = ¯ f|N .

(a) Replacing the ϕj we may assume that gj satisfies the conditions (i) and (ii) in place of f¯.

29 ¯ 2n n ϕj ◦ ψλj ◦ f(Vj) ⊂ C (1) ∩ R

n 2n R R

2n 2n C (xλ , 1) M n C (3) j U N · ψλj xλj 2n ¯ C (1) f(Vj ) O·

V j hj ¯ fj

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 −1 (gj|Nj ) (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. 2n Induction step. Now we construct gj+1 : M → R . Consider the map 0 −1 n 2n ϕj+1 ◦ gj ◦ (hj+1) : C (3) → C (1) . 2n n Define a projection π : R → R by π(x1, . . . , x2n) = (xn+1, . . . , x2n). Then by 0 the choice of ϕj+1 we have 0 −1 n −1 2n ϕj+1 ◦ gj ◦ (hj+1) (C (3)) ⊂ π (0,..., 0) = {(x1, . . . , xn, 0,..., 0) ∈ R } . And by Sard’s theorem the set of critical values of the smooth map

0 −1 n n π ◦ ϕj+1 ◦ gj :(U ∪i≤j Wi) ∩ gj (Uλj ) → C (1) ⊂ R n has measure zero. Hence, we may select a point cq ∈ R satisfying the following: (1) cq is sufficiently close to (0,..., 0). 0 (2) cq is a regular value of π ◦ ϕj+1 ◦ gj. 2n (3) If the points p1 and p2, p1 6= p2, satisfy gj(p1) = gj(p2) ∈ C (xλj , 1), then 0 0 −1 ϕj+1 ◦ gj(p1) = ϕj+1 ◦ gj(p2) ∈/ π (cq) .

30 2n Using the cq, we define a smooth map gj+1 : M → R by  −1 n gj(x) if x ∈ M \ hj+1(C (2)) , gj+1(x) = 0 −1 0 (ϕj+1) {ϕj+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 guarantee q+1 that gj+1 is an immersion and that gj+1 is a δ/2 -approximation of gj. Further- 0 0 0 more setting Nj+1 = Nj ∪W j and noting that Nj+1 is compact, we see that for any 0 −1 point p of g (N ), the set (g | 0 ) (p) contains at most two points, and j+1 j+1 j+1 Nj+1 if it contains two points, then gj+1 intersects transversely at p. It is also routine ¯ ¯ that gj+1|N = f|N . In addition, by readjusting the ϕj we see that gj+1 replacing f satisfies the conditions (i) and (ii). Finally, by examining the definition of gj+1 carefully we see that gj+1 and gj are regularly homotopic relative to N. 2n Defining a map g : M → R by

g(x) = lim gi(x) , i→∞ we have what we wanted.  In order to introduce additional regular self-intersections to a given mapping 2n M → R we need to construct a model of an immersion with a single regular 2 self-intersection. The map α : R → R defined by x 1 y = x − , z = (4.2) (1 + x2) (1 + x2) is a completely regular immersion, with one regular self-intersection at (0, 1/2), see Figure 4.3. Furthermore, there exists r > 0 so large that α is the identity map outside the unity D(0, r), x y = x − −→ x , (1 + x2) |x|→∞ 1 z = −→ 0 (1 + x2) |x|→∞ so that, for large enough x, we have approximately x 7→α (x, 0) .

n 2n For n ≥ 2, α : R → R is defined as follows. n α(x1, . . . , xn) = (y1, . . . , y2n), (x1, . . . , xn) ∈ R , (4.3) with 2x y = x − 1 , y = x , i = 2, . . . , n , 1 1 u i i and 1 x x y = , y = 1 i , i = 2, . . . , n , n+1 u n+i u

31 1 2 Figure 4.3. The completely regular immersion α : R → R de- fined by (4.2) with a single (regular) self-intersection at (0, 1/2). where 2 2 u = (1 + x1) ··· (1 + xn) . The rank of a mapping is independent of the choice of coordinate maps for the n 2n charts in the definition of rank. Since we are mapping R into R we can choose the identity maps as coordinate maps. With this, the rank of α is simply the rank of its Jacobian matrix (Dα)(x) which is

2  2(1−x1) 4x1x2 4x1xn  1 − u(1+x2) u(1+x2) ... u(1+x2 )  1 2 n   0 1 0 0     0 0 ... 0     . .   . .     0 0 ... 1  .  −2x1 −2x2 −2xn   2 2 ... 2  u(1+x1) u(1+x2) u(1+xn)  2 2   x2(1−x1) x1(1−x2) −2x1x2xn   u(1+x2) u(1+x2) ... u(1+x2 )   1 2 n   . . .   . . .   2 2  xn(1−x1) −2x1x2xn x1(1−xn) 2 2 ... 2 u(1+x1) u(1+x2) u(1+xn) 2 Not all elements of the first column are zero. If x1 6= 0 then −2x1/[u(1 + x1)] 6= 0. If x1 = 0 then the first element is 1 − 2/u. If this element is zero then u = 2 which means that some of the xi’s, i = 2, . . . , n, must be nonzero. Then the 2 2 corresponding term xi(1 − x1)/[u(1 + x1)] = xi/u is nonzero. Thus, with the 1’s in the other columns, there are n linearly independent nonzero columns, i.e. the rank n 2n of D(α : R → R )(x) is n and α is an immersion. We now want to show that α has exactly one (regular) self-intersection, i.e. a double point which intersects transversely. Let 0 0 0 x = (x1, . . . , xn), x = (x1, . . . , xn) , 0 0 2 0 2 u = (1 + (x1) ) ··· (1 + (xn) ) , 0 0 0 α(x) = (y1, . . . , y2n), α(x ) = (y1, . . . , y2n) . 0 0 0 At a double point α(x) = α(x ) we have x 6= x . First we observe that xi = xi for 0 0 i = 2, . . . , n by the definition of α. That x 6= x then implies that x1 6= x1. Secondly,

32 0 0 since yn+1 = yn+1 we must have u = u . Combining this with the first observation 2 0 2 0 we have that 1 + x1 = 1 + (x1) , which is only possible if x1 = −x1 6= 0. With this, 0 2 yn+i = yn+i can only be true if xi = 0, i = 2, . . . , n. Therefore, u = 1 + x1 and 0 y1 = y1 tells us that 2x 2x x − 1 = −x + 1 1 u 1 u so 2 2 1 − = −1 + u u thus 2 u = 2 = 1 + x1 or

x1 = ±1 . The only double point turns out to be y = α(±1, 0,..., 0) .

Set x+ = (1, 0,..., 0) and x− = (−1, 0,..., 0). For α to intersect transversely we need, by the discussion preceding the definition 4.3, ( ) h i n n m m m Dα(x+) Dα(x−) v v ∈ T × T = {since T = , x ∈ } = x+ R x− R xR R R

( ) h i 2n 2n = Dα(x+) Dα(x−) v v ∈ R = R .

This is equivalent to the matrix [ Df(p1) Df(p2) ] being invertible,  1 0 ... 0 1 0 ... 0   0 1 ... 0 0 1 ... 0     ......   ......       0 0 ... 1 0 0 ... 1  Dα(x+) Dα(x−) =   −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 some definitions regarding the notion of orientation. Definition 4.6. Consider an n-dimensional manifold (M, D). Let D = [S ], S = −1 {(Vj, ϕj)}. For x ∈ Vi ∩ Vj let aji(x) be the Jacobian matrix of ϕj ◦ ϕi at ϕi(x): −1 aji(x) = D(ϕj ◦ ϕi )(ϕi(x)), x ∈ Vi ∩ Vj .

33 Then it is easy to see that

akj(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) and we 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 take this to be the definition of orientable manifold. For the special case of Euclidean spaces, which are all orientable, we also need the following. n Definition 4.7. Two ordered set of n vectors, B and B0, that are bases for R , are said to be equivalent if their transition matrix has positive determinant. This n partitions the set of all bases for R into two equivalence classes. If the transition n matrix from a basis B of R to the standard basis has positive determinant, then n B is said to define the positive orientation on R . Otherwise it is said to define the n negative orientation on R . 2n Definition 4.8. Let f : M → R 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 be ordered sets of linearly independent vectors, which define the orientations of TpM and TqM. We say that the self-intersection at f(p) has positive type or negative type depending on whether the ordered set of vectors 2n (df)p(u1),..., (df)p(un), (df)q(v1),..., (df)q(vn) ∈ Tf(p)R 2n defines the positive orientation or the negative orientation in R , , and we define the intersection number of this self-intersection to be +1 or −1 accordingly. The intersection number If of f is the sum of intersection numbers of self-intersections of f: If ∈ Z. (ii) The case where M is non-orientable or n is odd. In this case we define the intersection 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 2n a completely regular immersion f : M → R with If = m.

(ii) If M is non-orientable or n is odd, then for any m ∈ Z2, there exists a 2n completely regular immersion f : M → R with If = m. 2n Proof. By Theorem 4.5 there exists a completely regular immersion f0 of M in R . Take a point x0 of M and some neighbourhood U of x0 to replace f0 by the map β with exactly one self-intersection defined in (4.3) or by the composite of β with 2n 2n the map r : R → R 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 will be introduced, including a triple point or even a quadruple point. The intersection number of the new completely regular immersion f is If0 + 1 or If0 − 1. We repeat this 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 of B we mean a continuous map s : X → B with p ◦ s = idX , the identity on X.A smooth map s which is a section of a smooth coordinate bundle is called a smooth section. m Lemma 4.11. Let B = {B, p, |K|, R ,O(m)} be an m-vector bundle over a poly- 0 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- 0 0 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 dim K ≤ m − i, we can 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 together with the necessary background in theory. m Fix a point p of |K| and let Rp denote the fiber over p. Then the desired ζi(p) may be a unit vector in the orthogonal complement of the (i − 1)-subspace spanned m−i ⊥ by ζ1(p), . . . , ζi−1(p), i.e., we want ζi(p) ∈ Sp ⊂ {{ζ1(p), . . . , ζi−1(p)}} ⊂ m m−i Rp , where Sp 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 reduces to the problem of extending a cross section over |K0| to |K| in the (m − i)-sphere 1 bundle over |K|.  For Lemma 4.12 and Lemma 4.13, consider a completely regular immersion f : 2n M → R (see Figure 4.4 but keep in mind that the Whitney Trick is really only valid for manifolds M of dimension 3 or higher). Assume f has two regular self- intersection points q and q0 of opposite sign,

f(p1) = f(p2) = q, p1 6= p2 ; 0 0 0 0 0 f(p1) = f(p2) = q , p1 6= p2 . 0 Let C1 and C2 be disjoint in M; Ci connecting pi and pi without passing through any other point where f has a self-intersection, i = 1, 2. Let Bi be the 0 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 interval

A1 = {(x, 0) | x ∈ [0, 1]} , √ and A2 be the part of the circle of radius 1, centered at (1/2, − 3/2), connecting 0 the 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 C1 0 p1 p1 M 1 n M 0 M2 p2

p2 C2

f(M n) f

f(M2)

B1 = f(C1) f(M1) q q0

B2 = f(C2)

ψ(τ 00) = σ2

ψ τ 00 x2

τ 0 A2

r A1 0 · ·r x1

Figure 4.4. (Upper two, with the mapping f.) The completely 2n regular immersion f : M n → R considered in Lemma 4.12, Lemma 4.13, and Theorem 4.14. (Lower two, with the mapping 2 ψ.) Construction in R for the Whitney trick

0 2 (See Figure 4.4.) Let A = A1 ∪ A2. Let τ be a small neighbourhood of A in R and let τ 00 be the region enclosed by A. Lastly, set τ = τ 0 ∪ τ 00. With this there is an 2n embedding ψ : τ → R satisfying the following: 0 0 (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. 2n Proof. We prove the theorem for an embedding ψ : τ 0 → R to begin with.

36 Change f slightly (by so little that it is still an embedding) at pi in Mi so that f maps a neighbourhood of pi in Mi surjectively on the exponential image of some neighbourhood of q in n Ti := Tqf(Mi) .

Near q we assume that f maps a neighbourhood of Ci in Mi onto the exponential n image of some neighbourhood of a line in Ti , given by its intersection with 2 T := T1f(M1) × Tqf(M2) . 1 Call the lines Ti , i.e. 1 2 Ti := T ∩ Tqf(M) . Now we define the linear mapping φ : V¯ → T 2 which maps a closed neighbourhood V¯ of r in τ 0 to a closed neighbourhood of q in T 2. Then the image of the restriction φ| is a closed neighbourhood of q in T 1. In a similar way we define the linear V¯ ∩Ai i mapping φ0 of a closed neighbourhood V¯ 0 of r0. Now we parametrize Ai, 2 ri : [0, 1] → Ai ∈ R , so that φ(ri(t)) = qi(t), wherever φ is defined, meaning that these qi(t) are points 1 on the part of Ci mapped onto the exponential image of Ti . Below we will, among other things, extend V¯ and V¯ 0 over all of A in order for φ to fulfill the rest of (I) together with φ0. 0 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) ∈ Tr0 A2, u1(1) is pointed backward along A2, ( (4.5) u2(0) ∈ TrA1, u2(0) is pointed forward along A1,

u2(1) ∈ Tr0 A1, u2(1) is pointed backward along A1.

(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; (4.6) u2(t) rotates clockwise as t goes from 0 to 1. (See Figure 4.5.)

37 x2

A 2 u2(t)

r u1(t) 0 • •r x1 A1

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

¯ 2 1 Currently, φ(V ) lies in T and contains part of Ti which constitutes part of 1 f(Ci) = Bi. Further away from q, Bi does not, necessarily, coincide with Ti . Since ¯ 1 we want φ to satisfy (I), we extend V in such a way that even if qi(t) 6∈ Ti we still have qi(t) ∈ φ(V¯ ). This is done as follows. Let Ri(t) be a line segment collinear with ui(t) and centered at ri(t). Assume that part of the boundaries of both V¯ and 0 V¯ are formed by the Ri(t)’s, whose length ρ is such that no R1(t) touches an R2(t) 0 if r1(t), r2(t) 6∈ V¯ ∪ V¯ . (See Figure 4.6.) ¯ ¯ 0 Now define the vector field vi(t) = (dφ)ri(t)(ui(t)) for ri(t) ∈ [V ∪ V ] ∪ Ai. We 0 have that vi(t) is in f(V¯ ∩ V¯ ) and by Lemma 4.11 we can extend it to Bi without making it touch f(M) at qi(t). For linear maps, the Jacobian matrix is constant and equal to the coefficient matrix. Thus, on V¯ ∪ V¯ 0 where φ is linear, we have

vi(t) = (dφ)ri(t)(ui(t)) = Dφ(ri(t)) · ui(t) = φ(ui(t)) . 0 Therefore, for ri(t) ∈ V¯ ∪ V¯ , we have 0 φ(ri(t) + αui(t)) = qi(t) + αvi(t), ri(t) ∈ V¯ ∪ V¯ , |α| ≤ ρ .

We can choose closed intervals Ii(t) ⊂ [−ρ, ρ] such that 0 {ri(t) + αui(t) | t ∈ [0, 1], i = 1, 2, α ∈ Ii(t)} =τ ¯ , and hence extend φ to the closed neighbourhoodτ ¯0 of A. Now, we can extend φ continuously over τ as ψ0, which is a smooth map by a smooth approximation. We find the embedding ψ as a δ-approximation of ψ0 by Theorem 3.22, since 2n ≥ 5. The embedding ψ satisfies (II), and since 2 + n < 2n it also satisfies (III).  This last condition is important as a property of the 2-cell σ2 = ψ(τ), which we will need in Theorem 4.14. We will also need a certain set of sections on the tangent 2n bundle T R restricted to σ2. This is taken care of in Lemma 4.13. Lemma 4.13. Assume q and q0 are self-intersections of f of opposite types. Then 2n there exist sections w1, . . . , w2n in T (R )|σ2 , the restriction of the tangent bundle 2n 2n T (R ) of R to σ2, satisfying the following: ∗ 2 ∗ ∗ (0) For each q ∈ σ , w1(q ), . . . , w2n(q ) are linearly independent.

38 x2

r 0 • •r x1

Figure 4.6. Modifying the neighbourhoods V¯ and V¯0.

(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 n−1 n−1 V1 = {{e3, . . . , en+1}},V2 = {{en+2, . . . e2n}} , 2n where {{... }} denotes the subspace of R spanned by {... }. For each point q∗ of B1 put n−1 ∗ V1 (q ) = {v ∈ Tq∗ f(M1) | v⊥Tq∗ B1} . S n−1 ∗ Setting = ∗ V (q ), we obtain an (n − 1)-dimensional vector bundle B1 q ∈B1 1 B1 = {B1, π1,B1} over B1. As the base space B1 is contractible, the bundle B1 is trivial. 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 n−1 ∗ V2 (q ) = {v ∈ Tq∗ f(M2) | v⊥Tq∗ B2} , we obtain an (n − 1)-dimensional vector bundle B2 = {B2, π2,B2}, where B2 = S n−1 ∗ 0 ∗ V (q ). Again there exist linearly independent vector fields w , w ,..., q ∈B2 2 2 n+2 ∗ w2n in B2, which define an orientation of f(M2) at each point q of B2. Here we 0 ∗ assume w2(q ) to be an element of Tq∗ B2, which points in the positive direction along B2. 2 2n 2 2 Let B = {σ × R , p1, σ } be a 2n-dimensional trivial bundle over σ . The re- striction B|B2 of B over B2 has n+1 linearly independent sections w1, w2, wn+2,..., ∗ ∗ 0 ∗ ∗ 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 q0 have opposite types. By suitable choices 0 of w1, w3, . . . , wn+1 in B1 and w2, wn+2, . . . , w2n in B2 we may assume that for q∗ = q or q∗ = q0, the vectors

∗ ∗ ∗ 0 ∗ ∗ ∗ w1(q ), w3(q ), . . . , wn+1(q ), w2(q ), wn+2(q ), . . . , w2n(q )

2n define the orientation opposite from the preassigned orientation in R . 0 0 0 0 2 Now we can deform w2(q) and w2(q ) to w2(q) and w2(q ) keeping them in T (σ ) and out of T (B1); thus, these vectors remain linearly independent of the above vectors. Hence, the vectors

∗ ∗ ∗ w1(q ), w2(q ), . . . , w2n(q )

2n define the positive orientation of R at q∗ = q or q∗ = q0. Therefore, we can extend the section wn+1 over B2 while keeping its linear independence with the other vectors. 2 Using Lemma 4.11 again we extend the sections w3, . . . , wn+1 over σ so that 2 w1, . . . , wn+1 are linearly independent at each point of σ . 2 Finally, we extend the sections wn+2, . . . , w2n defined over B2 to σ while keeping their linear independence. This is possible since we may think of B2 as a smooth 2 deformation retract of σ . Hence, we have shown the lemma is valid.  As a last step before proving Whitney’s second embedding theorem, we now show how to “add” and “remove” pairs of regular self-intersections. Thus we prove the existence of completely regular immersions with one pair more or less than a given completely regular immersion f, of Figure 4.4. The following theorem states that it is, under certain circumstances, possible to both add and remove pairs of regular singlurarities. For the second embedding theorem, we only need to be able to remove pairs of regular singularities. Therefore, only a scetch is made of how to add them.

2n Theorem 4.14. Let f : M → R be an immersion, where M is an n-dimensional manifold and n ≥ 3. Then there exists a regular homotopy {ft|t ∈ [0, 1]} of f satisfying 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 one self-intersection, then there exists a regular homotopy {ft} of f such that the number of self-intersections of f1 is two less than that of f = f0.

40 x2 x2

p D1 • • • • p q x1 x1 x2

• q x1

• p

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

Proof. Notice first that by Theorem 4.5 we may assume that f is a completely 2n regular immersion. Take a sufficiently small n-disk Dn in f(M), Dn ⊂ f(M) ⊂ R , 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 D1 2 in R as shown in Figure 4.7 and make two self-intersections by moving the disk 1 2 around by a regular homotopy, say, q = 0, p ∈ D1 ⊂ R ⊂ R . n 2n (b) Assume Dn is embedded in R ⊂ R . Let q = 0 and p be points of Dn with p 6= q. Assume p sits on the x1-axis. Pull up p to position some neighbourhood of x parallel 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 on the x1-axis; we have created two self intersections. We show how to eliminate two self-intersections in case (ii) where either M is non-orientable or n is odd, and how to eliminate two self-intersections of distinct types in case (i) where M is orientable and 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). 2 Next, deform f through σ 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. 2 Now, for the case of removing pair of regular singularities. Consider τ ⊂ R ⊂ 2n 2n ∗ R . For each point r = (a1, a2, . . . , a2n) of R set r = (a1, a2, 0,..., 0) and

2n ! 2n ∗ X ∗ X ∗ ψ(r) = ψ r + aiei = ψ(r ) + aiwi(ψ(r )) . i=3 i=3

41 x2

x2 = µ(x1)

A2

• • r 0 x1 A1 r

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

∗ 2 ∗ ∗ For each point q of σ , the vectors w1(q ), . . . , w2n(q ) are linearly independent and they are smooth in q∗. Therefore, ψ when considered as a map from a neigh- 2n bourhood of σ2 to R has the nonzero Jacobian matrix at each point in its domain. Hence, we have the inverse ψ−1. Setting

−1 −1 N1 = ψ (f(M1)),N2 = ψ (f(M2)) ,

2n we see that N1 and N2 are contained in a neighbourhood U of τ in R . If we can deform N2 in U so N2 does not intersect N1 (i.e., there exists an isotopy {it : t ∈ [0, 1]} of the i : N2 → U such that i0 = i, i1(N2) ∩ N1 = ∅), the family {ψ ◦ it} defines a deformation of f and the number of self-intersections of 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 given ∗ above for r ∈ A1, Tr∗ N1 is in the (x1, x3, . . . , xn+1)-plane. Hence Tr∗ π(N1) is also in 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 A2 0 and 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 from 1 the (x1, x2)-plane are each less than ε. Consider a smooth function ν : R → R as follows: |ν(λ)| ≤ 1, ν(0) = 1, ν(λ) = 0 if |λ| ≥ ε2 .

2n 2n Now define a map θt : R → R by 2 2 2n θt(r) = r − tν(x3 + ··· + x2n)µ(x1)e2, r = (x1, . . . , x2n) ∈ R . 2n By the definition of ν, θt is the identity map outside N2. Evidently the {θt : R → 2n R } is a regular homotopy with θ0 = 1. When t = 1, θ1 maps the portion of N2 on A2 to the halfplane x1 < 0. But π(θt(N2)) = π(N2) and π(N1) ∩ π(N2) is on the x1-axis, and so N1 ∩ θ1(N2) must be on the x1-axis. However, θ1(N2) does not intersect the x1-axis, and so θ1(N2) must be empty.

42 Furthermore, since ψ(τ)∩f(M) = B no new self-intersection will arise if we take ε sufficiently small. This concludes the proof for the case that M is orientable and n is even. In other cases we can also remove pairs of self-intersections by regular in just about the same way as above. To do this we only need to show that C1 and 0 C2 can be chosen so that q and q have opposite types.

0 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 q0 is of the same 0 0 0 type, we choose a curve C2 from p2 to p2 so that C2 ∪ C2 reverses the orientation 0 0 in M. Then q and q are of opposite type with respect to C1 and C2.

The case n is odd and M is orientable. Assume q and q0 have the same type for 0 0 0 0 C1 and C2, Then choose a curve C1 from p1 to p2 and a curve C2 from p2 to p1 as 0 follows. The curve Ci agrees with Ci near the starting point and with Cj, i 6= j, 0 0 near the endpoint. The neighbourhood Mi of Ci is chosen in such a way that Mi 0 0 0 0 agrees with Mi near the point pi and with Mj near pi, j 6= i. Orient Mi and Mi 0 0 with the preassigned orientations near pi and pi. Then q and q have the opposing 0 0 orientations with respect to (M1,M2).  Immersions and embeddings of non-compact manifolds are proper, in addition to what the original Definition 2.11 specifies. Continuous maps are proper if the inverse image of every compact set is compact. With this, we finally have everything we need for the proof of Whitney’s second embedding theorem. Theorem 4.15 (Whitney’s second embedding theorem). Every n-dimensional dif- 2n ferentiable manifold can be embedded in R . Proof. Let M be an n-dimensional manifold. The theorem is routine for n = 1 as M 1 is a finite union of S1. Case n = 2. We can embed the sphere S2, the projective space RP 2 and the 4 K2 in R . By the classification theorem for closed surfaces, M is a connected sum of a finite number of copies of the above surfaces. Hence, we can 4 embed M in R (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 : 2n M → R with If0 = 0. By Theorem 4.14 we can remove all self-intersections. 2n Thus, we have obtained an embedding f1 : M → R as an injective immersion on a compact set. Now assume that M is not compact and let (pi, qi) denote the double points of f0. For each pi, take a curve γi from pi to infinity. Now define neighbourhoods Ui, for each curve γi, such that Ui intersects no other Uj or contains any other pj. n−2 The points of the Ui’s may be written as (p, r, s) with p ∈ S0 , 0 ≤ r ≤ 1 and 0 ≤ s < 1, such that r = 0 and 2δ ≤ s < 1 represents the curve while r = 1 or s = 0 gives the boundary of Ui. Next, define the deformation  φt(p, r, s) = p, r, [1 − (1 − δ)(1 − r)t]s ,

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

44 5. Acknowledgements For helping me with subtle but important mathematical details, the art of writing a beautiful essay, and for giving me moral support, I wish to thank my supervisor Per Ahag,˚ he who mysteriously combines godlike mathematical and socoeu ail skills. I also wish to thank Robin Ekman, Klara Leffler and my examinator Lisa Hed, for reading my essay and giving me invaluable constructive feedback.

45

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] Dieudonn´eJ., A history of algebraic and differential topology 1900-1960. Modern Birkh¨auser Classics. Birkh¨auserBoston, Inc., , MA, 2009. [3] Donaldson S. K., An application of to four-dimensional topology. J. Differential Geom. 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] L¨utzenJ., 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 University Press of Virginia, Charlottesville. Princeton, New Jersey, 1965. [12] Prasolov V. V., Elements of combinatorial and differential topology. Translated from the 2004 Russian original by Olga Sipacheva. Graduate Studies in Mathematics, 74. American Mathematical Society, Providence, RI, 2006. [13] Prasolov V. V., Elements of homology theory. Translated from the 2005 Russian original by 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 Landmarks in Mathematics. Princeton Paperbacks. 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: [email protected]

47