Differential Topology

Home , Differential topology, Local diffeomorphism, Submersion (mathematics)

Diﬀerential Topology

Andrew Kobin Fall 2015 Contents Contents

Contents

0 Introduction 1 0.1 Review of General Topology ...... 1

1 Smooth Manifolds 4 1.1 Smooth Maps and the Tangent Space ...... 5 1.2 Submanifolds ...... 7 1.3 Abstract Manifolds ...... 8 1.4 Manifolds With Boundary ...... 13

2 Regular Values 15 2.1 Regular and Critical Values ...... 15 2.2 Sard’s Theorem ...... 22 2.3 Homotopy and Degree mod 2 ...... 24 2.4 Orientation ...... 27 2.5 Brouwer Degree ...... 28

3 Transversality and Embeddings 34 3.1 Transversality, Immersion and Submersion ...... 34 3.2 Embedding Theorems ...... 38

4 Vector Bundles 41 4.1 Vector and Tangent Bundles ...... 41 4.2 Sections ...... 44 4.3 Constructing New Vector Bundles ...... 47

i 0 Introduction

0 Introduction

These notes cover a course in differential topology taught by Dr. Slava Krushkal in the fall of 2015 at the University of Virginia. The main references for the course are Wallace’s Differ- ential Topology: First Steps and Milnor’s Princeton lectures, Topology from the differentiable viewpoint. The topics covered in the course are: Smooth maps and manifolds

Tangent spaces

Submanifolds of Rn and the embedding theorem Transversality

Regular and critical points

Sard’s theorem and Brouwer’s ﬁxed point theorem

Degree of maps

Vector ﬁelds and the Euler characteristic

An introduction to vector bundles and diﬀerential forms.

0.1 Review of General Topology

The formal objects in general topology were only laid out about a century ago, but the concepts they represent go back much further in time. Deﬁnition. A topology T on a set X is a collection of subsets of X which satisfy

(1) ∅,X ∈ T . S (2) For every arbitrary collection Uα ∈ T , α Uα ∈ T . Tk (3) For every ﬁnite collection U1,...,Uk ∈ T , i=1 Ui ∈ T . The pair (X, T ) is called a topological space, although this is often abbreviated by X when the topology is understood. One of the most important topologies is the one generated by a metric.

Deﬁnition. A metric on a space X is a function d : X × X → R such that (1) (Positive semideﬁnite) d(x, y) ≥ 0 for all x, y ∈ X and d(x, y) = 0 if and only if x = y.

(2) (Symmetric) d(x, y) = d(y, x) for all x, y ∈ X.

(3) (Triangle inequality) d(x, y) + d(y, z) ≤ d(x, z) for all x, y, z ∈ X.

1 0.1 Review of General Topology 0 Introduction

The sets that make up a topology are by convention called open sets or neighborhoods, while their complements are called closed sets.

Deﬁnition. A point p ∈ X is a limit point of a set A ⊂ X if every neighborhood U of p intersects A in a point other than p.

Deﬁnition. A point p ∈ X lies in the closure of A, denoted A, if every neighborhood U of p intersects A nontrivially.

It is easy to see from these two deﬁnitions that a set is closed if and only if it contains all of its limit points. For this reason it’s common to write the closure as A = A ∪ L, where L is the set of limit points of A. One of the most important conceits in mathematics is the idea of comparison. We generally accomplish this by constructing maps between objects (categorically, maps are called morphisms) that preserve a desired structure. In topology, the objects are spaces and the natural maps between them are continuous functions.

Deﬁnition. A function f : X → Y between topological spaces is continuous at a point p ∈ X if for every open set U containing f(p), f −1(U) ⊂ X is also open. The function is continuous (on X) if it is continuous at every point in X.

Deﬁnition. A continuous map f : X → Y is called a homeomorphism if it is one-to-one and onto (a bijection) and has a continuous inverse. In this case, X and Y are said to be homeomorphic, denoted X ∼= Y . One of the principal themes in topology is the study of topological properties, or properties preserved under homeomorphism. We recall several key topological properties next.

Deﬁnition. Given a topological space X, a subset A ⊂ X is compact if every open cover of A has a ﬁnite subcover.

Proposition 0.1.1. Closed subspaces of compact spaces are compact.

The Separation Axioms are additional assumptions we make on a topological space to ensure there are “enough” open sets in X to do certain things. The ﬁrst two are listed below.

Deﬁnition. A space X is T1 if every point set is closed.

Deﬁnition. A space X is Hausdorﬀ (also called T2) if for every pair of distinct x, y ∈ X, there exist disjoint open sets U and V such that x ∈ U and y ∈ V .

Proposition 0.1.2. If X is Hausdorﬀ and A ⊂ X is compact, then A is closed.

The converse of this statement is not true in general, and even in nice spaces (metric spaces) there are additional conditions (e.g. boundedness) to imply compactness.

Deﬁnition. A topological space X is said to be disconnected provided there exist nonempty, disjoint open sets U and V whose union is X. If no such sets exist, we say X is connected.

2 0.1 Review of General Topology 0 Introduction

In this course, we will implicitly assume that every map is smooth, i.e. infinitely differentiable on an open subset U ⊂ Rn. However, we need to define the notion of a smooth map on any subset of Rn, not just the open ones.

Definition. For an arbitrary subset S ⊂ Rn, a function f : S → Rm is smooth on S if for every x ∈ S, there exists a neighborhood W ⊂ n containing x and an extension R f˜ : W → Rm which is smooth on W and satisfies f˜ = f. W ∩S In differential topology, the fundamental notion of equivalence is even stronger than homeomorphism.

Definition. A diffeomorphism is a map f : X → Y that is a homeomorphism (continuous and invertible, with continuous inverse) that is smooth and whose inverse is also smooth. In the case that f is a diffeomorphism, we say X and Y are diffeomorphic.

An important theorem going forward is the Inverse Function Theorem from analysis, which we recall here.

Theorem 0.1.3 (Inverse Function Theorem). Suppose f : U → Rn is a smooth map on an open set U ⊂ n, and for a point x ∈ U, assume the diﬀerential D f is non-singular, or R x equivalently, the Jacobian ∂fi (x) is invertible. Then there exists a neighborhood V ⊂ U ∂xj containing x such that f|V : V → f(V ) is a diﬀeomorphism.

3 1 Smooth Manifolds

1 Smooth Manifolds

The fundamental object in differential topology is the smooth manifold. Imprecisely, a manifold is a topological space which is locally Euclidean. In this section we will present two different definitions of a smooth manifold, but ultimately we will see that they are equivalent.

Deﬁnition. A smooth n-dimensional manifold is a subspace M ⊂ Rk, k ≥ n, such that for any point x ∈ M, there is a neighborhood U ⊂ M containing x which is diﬀeomorphic to an open subset of Rn. We will often write M n to convey that M is an n-dimensional manifold.

Example 1.0.4. The unit circle S1 ⊂ R2 is a smooth one-dimensional manifold.

( x )

Given a point x ∈ S1, there are many good choices of a diﬀeomorphism, including

Coordinate projection: f(x1, x2) = x1. Stereographic projection from N = (0, 1).

Polar coordinates: (cos θ, sin θ) 7→ θ for θ ∈ (0, 2π).

Each of these will be a diffeomorphism on a different neighborhood of x. However notice that there’s no global diffeomorphism S1 → R since these spaces aren’t homeomorphic to begin with.

n Deﬁnition. A collection of diﬀeomorphisms gx : Ux → Vx ⊂ R , one for each point x ∈ M, is called an atlas of M. Each Ux is called a coordinate chart (or coordinate system) −1 around x, and the inverse map gx is called a parametrization about x. Note that when dim M ≥ 1, around any point x ∈ M, there are uncountably many choices of parametrizations and coordinate systems.

4 1.1 Smooth Maps and the Tangent Space 1 Smooth Manifolds

1.1 Smooth Maps and the Tangent Space

Loosely, the tangent space to a manifold M at a point x ∈ M is the set of all tangent vectors to M at x. This generalizes the concepts of tangent lines and tangent planes from multivariable calculus:

tangent plane M

tangent line f(x)

Deﬁnition. The tangent space to a smooth manifold M at a point x ∈ M is the set of tangent vectors to all parametrized diﬀerentiable curves lying in M and passing through x:

0 d TxM := α (0) = dt α(t)|t=0 α :(−ε, ε) → M is smooth and α(0) = x . Remarks.

α 1 A curve α :(−ε, ε) → M is smooth if the composition (−ε, ε) −→ M,→ Rk is smooth, which in turn means the composition with every coordinate projection is smooth:

α mi (−ε, ε) −→ M −→ R.

2 It is not immediately clear that TxM is closed under addition – however, scalar multi- ples are easily obtained by reparametrization of the curve.

Suppose M m and N n are smooth manifolds, possibly of diﬀerent dimensions. What does it mean for a map f : M → N to be smooth?

Deﬁnition. If M ⊂ Rk and N ⊂ R`, then f is smooth if for every x ∈ M there is an open k ˜ ` ˜ neighborhood W ⊂ R of x and a smooth map f : W → R such that f|W ∩M = f|W ∩M . Alternatively, if x ∈ M we have neighborhoods x ∈ U ⊂ M and f(x) ∈ V ⊂ N, and diﬀeomorphisms g : U → g(U) ⊂ Rm and h : V → h(V ) ⊂ Rn. Then f : M → N is smooth if for every x ∈ M, the composition h ◦ f ◦ g−1 is smooth:

5 1.1 Smooth Maps and the Tangent Space 1 Smooth Manifolds

f M N ∪ ∪ U V

g g−1 h

g(U) h(V ) ∩ ∩ h ◦ f ◦ g−1 Rm Rn These two notions of smooth maps between manifolds are equivalent. One may be more useful than the other in various circumstances. If f : M m → N n is a smooth map on manifolds, there is an induced map on the tangent spaces at any point x ∈ M: TxM → Tf(x)N. This map is linear, which we will justify in a moment. Explicitly, given v ∈ TxM, there is a curve α :(−ε, ε) → M such that α(0) = x and α0(0) = v. Then we define 0 Dxf(v) = (f ◦ α) (0). There is an alternate definition of the tangent space that relies less on the extrinsic choices of smooth curves. Choose a neighborhood U ⊂ M of x and a diffeomorphism g : U → g(U) ⊂ Rn. Then g−1 : g(U) → U ⊂ Rk has a well-defined differential, the Jacobian, so we may define the tangent space as the image of this differential:

−1 TxM = {Dg(x)g (y): y ∈ g(U)}.

Since the Jacobian is a linear map, it follows immediately that TxM is a vector space since it is the image of a linear map.

Lemma 1.1.1. The description of TxM is independent of the choice of U and g.

Proof. Suppose there are two neighborhoods U1 and U2 of x which have diﬀeomorphisms m m g1 : U1 → g1(U1) ⊂ R and g2 : U2 → g2(U2) ⊂ R . Then we must show −1 −1 im(Dg1(x)g1 ) = im(Dg2(x)g2 ).

Let U = U1 ∩U2; this is still a neighborhood of x. Restrict g1 and g2 to U and note that they are still diffeomorphisms, and g1(U) and g2(U) still contain g1(x) and g2(x), respectively. −1 The composition g2 ◦ g1 : g1(U) → g2(U) is a diffeomorphism and both g1(U) and g2(U) m −1 −1 −1 −1 −1 are open subsets of R . Thus we can write g1 = (g2 ◦ g2) ◦ g1 = g2 ◦ (g2 ◦ g1 ) and use the chain rule to differentiate:

−1 −1 −1 Dg1(x)g1 = Dg2(x)g2 ◦ Dg1(x)(g2 ◦ g1 )

−1 −1 But g2 ◦ g1 is a diﬀeomorphism, which means Dg1(x)(g2 ◦ g1 ) must be a vector space −1 −1 isomorphism. Therefore Dg1(x)g1 = Dg2 g2 as vector spaces.

6 1.2 Submanifolds 1 Smooth Manifolds

Proposition 1.1.2. The two deﬁnitions of tangent space are equivalent. Explicitly, the −1 image of the diﬀerential Dg(x)g is precisely the set of tangent vectors to smooth curves in M passing through x.

Proof. We first show that every vector v ∈ TxM is tangent to a smooth curve in M. Assuming m −1 v ∈ TxM, there exists a w ∈ R such that Dg(x)g (w) = v. Consider the curve β :(−ε, ε) → g(U) defined by β(t) = g(x) + tw. Then β(0) = g(x) and β0(0) = w. Define α = g−1 ◦ β which is now a smooth curve in M. It satisfies α(0) = x and, by the chain rule,

0 −1 0 −1 α (0) = (Dg(x)g ) ◦ β (0) = (Dg(x)g )(w) = v.

Hence v is the tangent vector to α at x. Conversely, take a curve γ :(−ε, ε) → M satisfying γ(0) = x, and denote γ0(0) = u. Then the composition g ◦ γ :(−ε, ε) → g(U) is a smooth curve in Rm. Set w = (g ◦ γ)0(0) as the tangent vector to g ◦ γ at g(x) in Rm. The chain rule gives us

−1 −1 −1 0 0 Dg(x)g (w) = Dg(x)(g ◦ g ◦ γ)(0) = (Dg(x)(g ◦ g)) ◦ γ (0) = γ (0) = u.

Therefore u ∈ TxM.

Lemma 1.1.3. For every x ∈ M, the dimension of TxM as a vector space is equal to the dimension of M as a smooth manifold.

Proof. Suppose dim M = m and take x ∈ M ⊆ Rk. Specify a neighborhood U ⊆ M of x and a diﬀeomorphism g : U → g(U) ⊆ Rm and its inverse g−1 : g(U) → Rk. Then Dg−1 is m k a linear map from a subset of R into R so its rank, which is the dimension of TxM, is at most m. On the other hand, look at g ◦ g−1 : Rm → Rm which is the identity. By the chain rule, −1 −1 −1 Id = Dg(x)(g ◦ g ) = (Dxg) ◦ (Dg(x)g ). Therefore the rank of Dg(x)g cannot be smaller than rank Id = m. Hence dim TxM = m. Remark. For each x ∈ M m, the tangent space is an m-dimensional vector space. However, for two diﬀerent points x, y ∈ M, TxM 6= TyM – they are formally disjoint. The collection of tangent spaces TxM over all points in M is what is known as a tangent bundle on M.

1.2 Submanifolds

Deﬁnition. A subspace S ⊂ M of a smooth manifold M is called a submanifold of M of dimension s if for every x ∈ S, there is an open set U ⊂ M containing x and a diﬀeomor- phism ϕ : U → ϕ(U) ⊆ Rm such that ϕ(S) = ϕ(U) ∩ Rs where

s m R = {(x1, x2, . . . , xs, 0, 0,..., 0) ∈ R : xi ∈ R}. Remark. It is immediate from the deﬁnition that submanifold is a manifold.

7 1.3 Abstract Manifolds 1 Smooth Manifolds

Example 1.2.1. Let M = R2 and consider the curve S = {(x2, x3) | x ∈ R}:

This curve has a cusp at (0, 0) so we should not expect S to be a smooth submanifold of R2. If we identify S as the graph of the function f(x) = (x2, x3), we see that f 0(0) = (0, 0) but this is not enough to conclude that S is not a submanifold. This demonstrates the diﬃculty in proving that a space is not a smooth manifold. If S were a smooth submanifold of R2, clearly it must be a 1-manifold (a 0-dimensional manifold would just be a collection of discrete points). Then each tangent space TpS for p ∈ S is a 1-dimensional vector space. Consider p = (0, 0), the problem point. If v ∈ TpS then v = α0(0) for some smooth curve α :(−ε, ε) → S with α(0) = (0, 0). Write α(t) = (x(t), y(t)). It is clear that y(t) has a local minimum at t = 0, so y0(0) = 0. At p = (0, 0), α is tangent to the y-axis so we must also have x0(0) = 0. To see this explicitly, diﬀerentiate the expression x(t)2 = y(t)3 twice and evaluate at t = 0:

2x(t)x0(t) = 3y(t)2y0(t) 2x0(t)2 + 2x(t)x00(t) = 6y(t)y0(t)2 + 3y(t)2y00(t) 2x0(0)2 + 2x(0)x00(0) = 6y(0)y0(0)2 + 3y(0)2y00(0) =⇒ 2x0(0)2 + 0 = 0 + 0 =⇒ x0(0) = 0.

0 0 0 Therefore α (0) = (x (0), y (0)) = (0, 0) but this contradicts the assumption that T(0,0)S is 1-dimensional. Hence S cannot be a smooth submanifold of R2.

1.3 Abstract Manifolds

The following example motivates a diﬀerent deﬁnition of a smooth manifold than the one introduced at the start of the chapter.

Example 1.3.1. The projective plane RP 2 (more generally, the projective n-space RP n for n ≥ 1) is defined in various ways. First, RP 2 can be defined as the quotient space S2/ ∼, where ∼ is the antipodal relation x ∼ −x. Alternatively, RP 2 can be identified with the set

8 1.3 Abstract Manifolds 1 Smooth Manifolds

of lines through the origin in R3. There is a third construction of RP 2 as a quotient space of a square [0, 1] × [0, 1] or a bigon:

The projective plane does not embed into R3 but it is an important example of a topological manifold so we would like to expand our definition of manifold to encompass cases like this. Definition. A topological manifold of dimension n is a Hausdorff topological space M

which admits a covering by countably many open sets {Ui}i∈N such that each Ui is homeo- n n morphic to an open subset of R , via ϕi : Ui → Vi ⊂ R .

Deﬁnition. M is a smooth manifold if, in addition, whenever two charts Ui,Uj intersect −1 n nontrivially, then the transition map ϕj ◦ ϕi : ϕi(Ui ∩ Uj) → R is a diﬀeomorphism.

M Ui x Uj −1 ϕ ϕi j

−1 ϕj ◦ ϕi

−1 −1 ϕi (Ui ∩ Uj) ϕj (Ui ∩ Uj)

Example 1.3.2. Again consider projective n-space P := RP n = Sn/ ∼, where x ∼ −x. 1 ∼ 1 (Note that in dimension 1, RP = S but this is the only uninteresting case.) Let’s show that RP n is a smooth manifold under our new definition. First, it’s easy to see that RP n is Hausdorff since it’s the quotient of a Hausdorff space (Sn). It is useful to define RP n in terms of homogeneous coordinates, as the following description makes clear: P = [x0, x1, . . . , xn]:[x0, x1, . . . , xn] = [tx0, tx1, . . . , txn] for any t ∈ R r {0} .

Then P may be covered by n + 1 charts, Vi = {[x0, x1, . . . , xn]: xi 6= 0}. These are all open subsets of P : the preimage of Vi under π is the union of two hemispheres, which are open in Sn, and projection maps are open; thus each V is open. The diﬀeomor- i x0 x1 xi−1 xi+1 xn phisms are ψi[x0, x1, . . . , xn] = , ,..., , ,..., . Since xi 6= 0 in Vi, each xi xi xi xi xi

9 1.3 Abstract Manifolds 1 Smooth Manifolds

−1 ψi is well-deﬁned and smooth (each one is rational). Inverses are given by ψi (u1, . . . , un) = [u1, . . . , ui−1, 1, ui, . . . , un], which are well-deﬁned since any point in Vi can be written x0 xn [x0, . . . , xi, . . . , xn] = ,..., 1,..., . xi xi

The inverses are smooth for the same reason that the ψi are smooth. Now for two charts Vi,Vj, any point such that xi, xj 6= 0 lies in their intersection. The −1 n transition map ψj ◦ ψi : ψi(Vi ∩ Vj) → R has the following formula:

−1 ψj ◦ ψi (u1, . . . , un) = ψj[u1, . . . , ui−1, 1, ui, . . . , un] u u u u 1 u u = 1 ,..., j−1 , j+1 ,..., i−1 , , i+1 ,..., n . uj uj uj uj uj uj uj

Since uj 6= 0 in ψi(Vi ∩ Vj), the transition map is easily seen to be a diﬀeomorphism of n ψi(Vi ∩ Vj). Therefore RP is a smooth manifold of dimension n. The next result gives us a way of constructing new manifolds out of existing ones.

Proposition 1.3.3. Let M and N be smooth manifolds of dimension m and n, respectively. Then M × N is a smooth manifold of dimension m + n.

Proof. If M and N are smooth manifolds, then for any distinct pairs of points (m1, n1) and (m2, n2) ∈ M × N, either m1 6= m2 or n1 6= n2 (or both). So without loss of generality suppose m1 6= m2; in this case there exist disjoint neighborhoods U, V ⊂ M with m1 ∈ U and m2 ⊂ V , since M is Hausdorff. Then U × N and V × N are disjoint open sets in M × N such that (m1, n1) ∈ U × N and (m2, n2) ∈ V × N. Therefore M × N is Hausdorff. m Next, suppose {Ui} is a countable basis for M with homeomorphisms ϕi : Ui → R n and {Vj} is a countable basis for N with homeomorphisms ψj : Vj → R . Then the sets Wij := {(x, y) | x ∈ Ui, y ∈ Vj} form a countable basis for M × N (finite products of m n ∼ m+n countable sets are countable). For each Wij, define a map ωij : Wij → R × R = R by (x, y) 7→ (ϕi(x), ψj(y)). Since each ϕi and ψj is a homeomorphism, ωij inherits continuity

(componentwise), surjectivity and injectivity. Therefore we have a countable basis {Wij}i,j∈N m+n of M × N and homeomorphisms ωij : Wij → R so M × N is a manifold. Finally, suppose Wij ∩ Wk` 6= ∅. Consider the restriction of ωij to Wij ∩ Wk`. Then we −1 m+n have a map ωk` ◦ ωij : ωij(Wij ∩ Wk`) → R which can be written

−1 ωk`(ωij (x1, . . . , xm,xm+1, . . . , xn)) −1 −1 −1 −1 = ωk` ϕi (x1), . . . , ϕi (xm), ψj (xm+1), . . . , ψj (xn) −1 −1 −1 −1 = ϕk(ϕi (x1)), . . . , ϕk(ϕi (xm)), ψ`(ψj (xm+1)), . . . , ψ`(ψj (xn)) .

−1 Since M is a smooth manifold, ϕk ◦ϕi is a diffeomorphism on Ui ∩Uk which contributes the −1 first m components of ωij(Wij ∩ Wk`). Likewise, N is smooth so ψ` ◦ ψj is a diffeomorphism −1 on Vj ∩ V`, which make up the last n components of ωij(Wij ∩ Wk`). Therefore ωk` ◦ ωij is componentwise a diffeomorphism, so it is a diffeomorphism. This proves that M × N is a smooth manifold of dimension dim M + dim N = m + n.

10 1.3 Abstract Manifolds 1 Smooth Manifolds

Corollary 1.3.4. The tangent space of M × N is T(x,y)(M × N) = TxM ⊕ TyN. Given our new deﬁnition, what should it mean for a map f : M m → N n to be smooth? The transition maps play a key role in answering this question. Let M have basis {Ui} and charts {ϕi} and let N have basis {Vj} and charts {ψj}. Deﬁnition. A function f : M → N between smooth manifolds is smooth at a point x ∈ M if for every choice of basis elements Ui ⊂ M containing x and Wk ⊂ N containing f(x), the ∗ −1 ∗ −1 −1 composition ψk ◦ f ◦ ϕi : Vi → ψ(Wk) is smooth, where f = f|f (Wk) : f (Wk) → Wk ⊂ N. Further, f is smooth if it is smooth at every point x ∈ M.

At the moment the above definition depends on the choices of Ui and Wk, but the condition that the transition maps in M and N are smooth ensures that our definition is well-defined. The previous definitions of a tangent space rely on the embedding of a manifold in Euclidean space. Here we give a generalization which works for an abstract manifold. For a manifold M, let C∞(M) be the vector space of smooth functions M → R.

Deﬁnition. A functional v : C∞(M) → R is called a derivation at a point x ∈ M if it satisﬁes the Leibnitz rule:

v(fg)(x) = v(f)g(x) + f(x)v(g) for all f, g ∈ C∞(M).

Deﬁnition. The tangent space to M at a point x ∈ M is the vector space of derivations at x: ∞ TxM = {v : C (M) → R | v is a derivation at x}.

Proposition 1.3.5. The deﬁnition of TxM in terms of point derivations is equivalent to the other deﬁnitions of tangent space.

Proof. We will equate the point derivations with tangent vectors to smooth curves. Let x ∈ M and set

0 T1 := {α (0) | α :(−ε, ε) → M is smooth with α(0) = x}; ∞ and T2 := {v : C (M) → R | v is a point derivation at x}.

Take a chart V containing x and a diﬀeomorphism ϕ : V → U ⊂ Rn. Let p = ϕ(x). We will prove the statement for p and transfer things back to the manifold with ϕ−1. First deﬁne

A : T1 −→ T2 d α0(0) 7−→ f 7→ (f ◦ α)| . dt t=0

d 0 The element dt (f ◦ α)|t=0 is really the directional derivative of f in the direction of α (0). Then for any f, g ∈ C∞(M), the chain and product rules show that

d(fg) df dg df dg (fg◦α)0(0) = ◦α0(0) = g + f ◦α0(0) = ◦ α0(0) g(0)+f(0) ◦ α0(0) . dt dt dt dt dt

11 1.3 Abstract Manifolds 1 Smooth Manifolds

This shows that the image A(α0(0)) is indeed a derivation at x. Conversely, if v is a derivation ∞ Pn at x then for any f ∈ C (M), we can write f(x) = i=1 ai(x)xi for some smooth functions ∂f ai(x) satisfying ai(0) = . Since v is linear, this implies ∂xi xi=0

n ! n X X v(f) = v ai(x)xi = v(ai(x)xi) i=1 i=1 n n X X = [v(ai)xi(0) + ai(0)v(xi)] = [v(ai) · 0 + ai(x)v(xi)] i=1 i=1 n X ∂f = (0)v(x ). ∂x i i=1 i Thus the expression of v(f) only depends on the partial derivatives of f, so we can deﬁne a map

B : T2 −→ T1

v 7−→ (v(x1), v(x2), . . . , v(xn)).

n 0 Then α(t) = (v(x1), v(x2), . . . , v(xn))t is a curve with α(0) = 0 ∈ U ⊂ R and α (t) = 0 B(v) ∈ T1. We now show that A and B are inverses. For α (0) = (w1, . . . , wn), denote the derivation A(α0(0)) by v. Then for f ∈ C∞(M),

n n d X ∂f dxi X ∂f v(f) = (f ◦ α)| = (0) (0) = (0)w . dt t=0 ∂x dt ∂x i i=1 i i=1 i

0 This implies B(v) = (w1, . . . , wn) = α (0). So A is one-to-one, but since dim T1 = dim T2, we have an isomorphism of vector spaces.

Example 1.3.6. We claim that GLn(R) is a smooth manifold. First, observe that Mn(R) is a 2 smooth manifold of dimension n , since every matrix (aij) ∈ Mn(R) can be viewed as a vector n2 (a11, a12, . . . , a1n, a21, . . . , ann). So as topological spaces, Mn(R) = R and consequently the space of n × n matrices is a smooth manifold under the global chart ϕ : Mn(R) → n2 R , ϕ(aij) = (a11, a12, . . . , ann). Consider the determinant map det : Mn(R) → R. Write (xij) as the matrix whose ijth entry is an indeterminate xij. Then det may be viewed (through cofactor expansion) as a polynomial in x11, x12, . . . , x1n, . . . , xnn and since polynomials are smooth, det is a smooth −1 map. The set U = Rr{0} is an open subset of the line, and by deﬁnition GLn(R) = det (U) so GLn(R) is an open subset of Mn(R). In general, open subsets of manifolds are quite easily seen to be submanifolds. In our setting, take ϕ from above and restrict it to the open set GLn(R) to obtain a chart on all of GLn(R). Moreover, since ϕ(GLn(R)) is an open subset n2 2 of R , we conclude that GLn(R) is a smooth manifold of dimension n .

12 1.4 Manifolds With Boundary 1 Smooth Manifolds

1.4 Manifolds With Boundary

We want to expand our notion of manifolds even further to include manifolds with boundary, such as the following ﬁgures.

1-dim. manifold 2-dim. manifold

2 2 3 x1 + x2 + x3 ≤ 1

3-dim. manifold

The idea we seek to describe is: M is a manifold with boundary of dimension n if at its n n boundary components, M is locally R+, the half-space R+ = {(x1, . . . , xn) | xn ≥ 0}. We will see shortly that the boundary components are all (n − 1)-manifolds.

Definition. A subset M ⊂ Rk is an n-dimensional smooth manifold with boundary if for every point x ∈ M, there is a neighborhood U of x which is diffeomorphic to an open n n n set V ⊆ R+, where R+ has the subspace topology inherited from R . The boundary of M, n ∼ n−1 denoted ∂M, consists of all points that are mapped to ∂R+ = {(x1, . . . , xn−1, 0)} = R under these diffeomorphisms.

Observe from this definition of ∂M that the boundary of an n-dimensional manifold M with boundary is a smooth (n − 1)-manifold – just take the restrictions of the charts to each boundary component and notice that they are diffeomorphisms between ∂M and n ∼ n−1 ∂R+ = R . Even better, ∂M is a smooth manifold without boundary so in other words, ∂2M = ∂(∂M) = ∅. This will allow us to define a chain complex on a manifold and compute its (co)homology.

Question. How do we define the tangent space TxM when x ∈ ∂M? Since ∂M is an (n − 1)-dimensional manifold, for each boundary point x ∈ ∂M, there is a subspace Tx(∂M) ⊂ TxM once we properly define TxM. In particular, for each x ∈ ∂M, n n there’s a neighborhood U ⊂ M and a diffeomorphism ϕ : U → V ∩ R+ where V ⊂ R is −1 n open. The inverse ϕ is defined on V ∩ R+ and is a diffeomorphism, so there exists an open n k −1 set V ⊆ and a smooth map Φ : V → such that Φ| n = ϕ . From this we define: e R e R V ∩R+

13 1.4 Manifolds With Boundary 1 Smooth Manifolds

Deﬁnition. For a boundary point x ∈ ∂M, the tangent space to M at x is the image of the diﬀerential of Φ, that is,

TxM = {Dϕ(x)Φ(y) | y ∈ Ve}.

We also need to extend the deﬁnition of a smooth map to maps between manifolds with boundary.

Deﬁnition. A map f : M → N between smooth manifolds with boundary is said to be smooth at a point x ∈ M in the usual way if x 6∈ ∂M. If x ∈ ∂M, f is smooth at x if for every choice of open sets x ∈ U ⊂ M and f(x) ∈ V ⊂ N and diﬀeomorphisms m n −1 −1 ϕ : U → ϕ(U) ⊂ R and ψ : V → ψ(V ) ⊂ R , there are extensions ϕe and ψe of ϕ −1 m n and ψ, respectively, such that the composition ψe ◦ f ◦ ϕe : W ⊂ R → Y ⊂ R on some neighborhoods W 3 x, Y 3 f(x) is smooth.

14 2 Regular Values

2 Regular Values

The degree of a map is highly related to regular values. First, we discuss the degree mod 2 of a map and later, after introducing orientability, we will generalize to integer-valued degrees. Our motivation for studying regular values is best seen in the following example. Consider a smooth map f : S1 → S1 from the unit circle into itself. The winding number of f is loosely deﬁned as ‘the number of times im f winds around, or covers the circle’. Consider the following functions.

f g id z y

The identity id : S1 → S1 wraps around the circle exactly once, so it is natural to expect the winding number ω(id) to be 1, and this is precisely the case. In contrast, consider a function f : S1 → S1 that winds 3/4 of the way around the circle, reversing directions once and then again to return to its starting point. Then f also has winding number 1. Finally, g wraps all the way around the circle once and then wraps all the way around again backwards; this corresponds to a winding number of 0. Our goal is to formalize this notion, by defining the degree (mod 2) of a map so that it is invariant under continuous deformations (homotopy). We first need to introduce the notion of a regular value, which in reality is one of the fundamental topics in differential topology.

2.1 Regular and Critical Values

We will start by studying regular values of maps between manifolds of the same dimension. Later this will be generalized.

Deﬁnition. Let f : M → N be a smooth map between two manifolds without boundary of dimension n. A point y ∈ N is called a regular value of f if for every x ∈ f −1(y), Dxf : TxM → TyN is a nonsingular mapping; that is, Dxf is a vector space isomorphism. If y is not a regular value of f, it is called a critical value.

Deﬁnition. A point x ∈ M is a regular point of f if Dxf is nonsingular. Otherwise x is called a critical point of f.

In the depiction of f above, the point y is not a regular value of f.

Deﬁnition. Let M,N be smooth, compact manifolds without boundary of the same dimension. For a smooth map f : M → N, the degree mod 2 of f is the cardinality mod 2 of −1 f (y) for any regular value y ∈ N, denoted deg2 f.

15 2.1 Regular and Critical Values 2 Regular Values

In the same picture of f, z is a regular value of f and one can see from the way im f is drawn that |f −1(z)| = 3 ≡ 1 (mod 2). So the degree mod 2 of f in the above picture is 1, which is the same as its winding number. Two questions immediately arise: (1) Do regular values exist for every function f? (2) Is |f −1(y)| mod 2 independent of the choice of regular value y of f? Sard’s Theorem (Section 2.2) will show that the answer to (1) is yes. In Section 2.3 we will prove that deg2 f is indeed well-defined. Let f : M → N be a smooth map between compact manifolds of equal dimension, x ∈ M with neighborhood U and diffeomorphism ϕ : U → Rn, and y = f(x) ∈ N with neighborhood V and diffeomorphism ψ : V → Rn. A direct consequence of the inverse function theorem is that if x is a regular point of f, then ϕ(x) is a regular point of the composition ψ ◦f ◦ϕ−1. In particular, by the inverse function theorem, there is a neighborhood W of ϕ(x) in Rn which is mapped diffeomorphically to a neighborhood of ψ(y) in Rn by ψ ◦ f ◦ ϕ−1. Therefore there is also a neighborhood, ϕ−1(W ), of x in M and a neighborhood Wf of f(x) in N such that f|W : W → Wf is a diffeomorphism. In broad terms, if x is a regular point, we have a good handle on what’s going on with our function (via a local diffeomorphism), but if x is critical, we likely have very little information on the behavior of f around x. Lemma 2.1.1. Let f : M → N be a smooth function between compact manifolds without boundary, such that dim M = dim N. Then the function |f −1(y)| is locally constant on regular values y ∈ N. Proof. Observe that since M and N are compact, f −1(y) is closed in M and for all x ∈ f −1(y)

there exists a neighborhood Ux such that f|Ux : Ux → Vx ⊂ N is a local diffeomorphism, for −1 some neighborhood Vx of y. First, suppose f (y) contains infinitely many points. Since M is −1 compact, there exists an infinite sequence x1, x2, x3,... contained in f (y) which converges to somex ¯ ∈ f −1(y) = f −1(y). Therefore by the inverse function theorem, there exists a neighborhood Ux¯ containingx ¯ such that Ux¯ → Vx is a diffeomorphism. So there cannot be −1 −1 any other values of f (y) in Ux¯, contradicting the convergence (xi) → x¯. Hence f (y) must be finite. −1 Now label the elements of f (y) = {x1, . . . , xn}. For each xi, there is a neighborhood

Uxi which is diffeomorphic to Vxi ⊂ N of y. Then V := Vx1 ∩ Vx2 ∩ · · · ∩ Vxn is an open set 0 −1 0 −1 containing y. If y ∈ V , then |f (y )| ≥ |f (y)|. Observe that y 6∈ f(M r (U1 ∪ · · · ∪ Un)). Then the set Ve := V r f(M r (U1 ∪ · · · ∪ Un)) is an open neighborhood of y in M on which |f −1(y0)| = |f −1(y)| for all y0 ∈ Ve. It follows from the proof that one can always find a neighborhood of a regular value of f consisting solely of regular values of f. This fact is not true when M is not compact. We now define regular values in the general case, when n 6= m. Definition. Suppose f : M m → N n is a smooth map between manifolds M and N, possibly of different dimensions. A point y ∈ N is called a regular value if for all x ∈ f −1(y), the differential Dxf : TxM → TyN is a surjection.

16 2.1 Regular and Critical Values 2 Regular Values

An immediate consequence of this deﬁnition is that if dim M < dim N, y ∈ N is a regular value of f if and only if y is not in the range of f. Theorem 2.1.2 (Regular Value Theorem). Let f : M m → N n be a map of smooth manifolds and m ≥ n. Then for every regular value y ∈ N, f −1(y) is a smooth submanifold of M of dimension m − n.

−1 Proof. Take x ∈ f (y). Since y is regular, Dxf : TxM → TyN is surjective. Then K = ker(Dxf) is a vector subspace of TxM of dimension m = n, by rank-nullity. Choose a linear m−n m−n map L : TxM → R such that L|K : K → R is an isomorphism of vector spaces (such a choice is possible since, for example, orthogonal projection onto K satisfies the desired properties for L). Define a new map F : M → N × Rm−n by F (x) = (f(x),L(x)). Then the two spaces have dimension m, so we can apply the inverse function theorem, but first we must show x is a regular point of F . Notice that DxF = (Dxf, DxL) = (Dxf, L) since L is linear. Then m−n the differential is a map TxM → TyN ⊕ R but ker(Dxf) is mapped nontrivially by L, so the differential at x must be nonsingular (in fact, it’s an isomorphism). Thus x is a regular point of F . Now the inverse function theorem says that there exists a neighborhood W of x in M m−n such that there is an open set Wf ⊂ N × R for which the restriction F |W : W → Wf is a diffeomorphism. It remains to show that F maps f −1(y) ∩ W diffeomorphically onto (y × Rm−n) ∩ Wf. Explicitly, F takes points in f −1(y) to points of the form (y, ∗) which lie clearly lie in y × Rm−n; and F maps W → Wf. Therefore F is a diffeomorphism from f −1(y)∩W to (y ×Rm−n)∩Wf. Hence f −1(y) is a submanifold of M of dimension m−n. Example 2.1.3. This theorem has many practical uses. For example, define f : Rn+1 → R Pn+1 2 −1 n+1 Pn+1 2 by f(x1, . . . , xn+1) = i=1 xi . Notice that f (1) = (x1, . . . , xn+1) ∈ R : i=1 xi = 1 = Sn, the unit n-sphere. If we can show 1 is a regular value, then the regular value theorem will imply Sn is a smooth manifold. n −1 n+1 Take x ∈ S = f (1) and consider the differential Dxf : TxR → T1R. Explicitly, ∂f ∂f this can be written Dxf = ,..., = (2x1,..., 2xn+1) but this is nonzero since ∂x1 ∂xn+1 at least one of the coordinates xi has to be nonzero for the point to sit on the unit n-sphere. Hence Dxf is surjective, so 1 is a regular point of f. This technique is a nice way to show certain objects like the n-sphere are manifolds.

2 Example 2.1.4. Recall from Example 1.3.6 that GLn(R) is a manifold of dimension n . We can use the regular value theorem to prove that SLn(R) is a submanifold of GLn(R). Again consider the determinant map det : Mn(R) → R. If we can show 1 is a regular −1 value of det, we’ll be done since by definition SLn(R) = det (1), so we can apply the regular 2 value theorem. For a matrix A ∈ SLn(R), the differential dA(det) is an n ×1 vector of partial ∂ det derivatives (A) . For a fixed xij, the partials with respect to xij are, up to sign, just the xij ij+1 n+1 cofactors. Indeed, det(xk`) = x11X11 − x12X12 + ... + (−1) xijXij + ... + (−1) xnnXnn, ∂ det ij+1 where Xij represents the ijth cofactor of X. Then (A) = (−1) Aij and the rest of xij the partials are the rest of the cofactors (with sign). In any case, since A ∈ SLn(R), A is invertible and therefore cannot have all cofactors equal to 0. Hence the differential is

17 2.1 Regular and Critical Values 2 Regular Values

2 non-singular, and since dA(det) is a map from a space of dimension n to one of dimension −1 1, it must in fact be surjective. Therefore 1 is a regular value of det, so det (1) = SLn(R) is a smooth submanifold of dimension n2 − 1.

Example 2.1.5. One can prove that other classical groups (subgroups of GLn(R)) are manifolds in a similar manner as in the last example. These are slightly more complicated to prove than SLn(R). For instance, consider O(n), the set of n × n matrices with orthonormal columns (equivalently, rows). Note that (AT A)T = AT (AT )T = AT A so this determines a map

f : Mn(R) −→ Symn(R) A 7−→ AT A,

T where Symn(R) = {C ∈ Mn(R) | C = C} is the space of symmetric n × n matrices. We know from linear algebra that Symn(R) is a vector subspace of Mn(R) and therefore is a n2 manifold when we view Mn(R) = R . It’s not too hard to show that Symn(R) is a manifold n(n+1) of dimension 2 . To deploy the regular value theorem, we will show that f is smooth, the −1 n × n identity matrix In is a regular value of f and f (In) = O(n). If A = (xij) ∈ Mn(R) then the ijth entry of f(A) = AT A is precisely

n n n X T X X [A ]ik[A]kj = [A]ki[A]kj = xkixkj. k=1 k=1 k=1

This shows that f is in fact a componentwise polynomial function in the entries xij of n2 A ∈ Mn(R), so identifying Symn(R) ⊂ R , we see that f is smooth. Moreover, this computation shows that

n T X f(A) = In ⇐⇒ A A = In ⇐⇒ xkixkj = δij for all 1 ≤ i ≤ m, 1 ≤ j ≤ n, k=1 Pn but the condition k=1 xkixkj = δij is precisely the orthonormality condition on the columns −1 of A. Hence we have shown that f (In) = O(n). Finally, we show that In is a regular value of f. Let A ∈ O(n) and view B ∈ TAMn(R) n2 ∼ n2 also as an n × n matrix since TAMn(R) = TAR = R . Then the diﬀerential dAf applied to B may be expressed as a directional derivative: f(A + tB) − f(A) (A + tB)T (A + tB) − AT A dAf(B) = lim = lim t→0 t t→0 t (AT + tBT )(A + tB) − AT A tBT A + tAT B + t2BT B = lim = lim t→0 t t→0 t = lim(BT A + AT B + tBT B) = BT A + AT B. t→0

To show this diﬀerential is surjective, it suﬃces to show that every matrix C ∈ Symn(R) ⊂ n2 T T R can be written as C = B A + A B for some B ∈ Mn(R). Given any n × n matrix C, 1 set B = 2 AC. Then

T T 1 T 1 T 1 T T 1 1 1 B A + A B = 2 (AC) A + 2 A AC = 2 C A A + 2 C = 2 C + 2 C = C.

18 2.1 Regular and Critical Values 2 Regular Values

−1 Thus dAf : Mn(R) → Symn(R) is surjective for every A ∈ f (In), so In is a regular value. n2 2 n(n+1) n(n−1) We conclude that O(n) is a submanifold of Mn(R) = R of dimension n − 2 = 2 . Corollary 2.1.6. For a smooth function f : M → N where dim M ≥ dim N, if y is a −1 −1 regular value of f then ker(Dxf) = Tx(f (y)) for every point x ∈ f (y).

−1 Proof. It’s easy to see that ker(Dxf) ⊇ Tx(f (y)) but both are (n − m)-dimensional vector spaces so they must be equal. To extend the regular value theorem to manifolds with boundary, we ﬁrst need the following results.

Lemma 2.1.7. Let f : M → N be a smooth map, where M is a manifold with boundary. −1 Suppose y ∈ N is a regular value both for f and for f|∂M . Let x ∈ f (y) ∩ ∂M. Then the −1 −1 tangent space Txf (y) at x to the submanifold f (y) is not contained in the tangent space Tx(∂M) to the boundary of M. Proof. Suppose dim M = m and dim N = n. By the existence of x ∈ f −1(y) ∩ ∂M, we may assume m ≥ n (otherwise the preimage of y under f is empty). We know that since y is −1 a regular value of f, dxf : TxM → TyN is surjective and linear, with ker dxf = Txf (y). Moreover, if dxf|Tx(∂M) : Tx(∂M) → TyN is also surjective, its kernel K has dimension −1 m − n − 1 by rank-nullity. Now if Txf (y) were contained in Tx(∂M), it would certainly be −1 contained in K = ker((dxf)|Tx(∂M)) as Txf (y) is the kernel of the entire diﬀerential. But −1 dim Txf (y) = m − n by the regular value theorem, so it cannot be contained in a vector space of dimension m − n − 1. Therefore we derive a contradiction from the assumption that −1 Txf (y) ⊂ Tx(∂M).

Lemma 2.1.8. Suppose g : M → R is a smooth map on a smooth manifold M with regular value y ∈ R. Then {x ∈ M | g(x) ≥ y} is a manifold with boundary g−1(y). Proof. If y is a regular value of g then 0 is a regular value of g(x) − y so we may assume y = 0 to begin with. We want to show that g−1([0, ∞)) is a manifold with boundary g−1(0). The regular value theorem for manifolds without boundary implies that since 0 is a regular value for g, g−1(0) is a manifold without boundary. Explicitly, g can be locally described −1 ∼ k−1 by g (0) = {0} × R . There was nothing special about 0 however, so for all y0 ∈ [0, ∞), −1 ∼ k−1 there is a local diﬀeomorphism g (y0) = {y0} × R . Patching these together gives us −1 ∼ k−1 g ([0, ∞)) = [0, ∞) × R . Theorem 2.1.9 (Regular Value Theorem (with Boundary)). Let f : M m → N n be a smooth map between manifolds M and N with boundary and suppose y ∈ N is a regular value both −1 for f and for f|∂M . Then f (y) is a smooth submanifold of M whose (possibly empty) boundary satisﬁes ∂(f −1(y)) = f −1(y) ∩ ∂M.

Proof. Let z ∈ f −1(y). Then either z ∈ Int(M) or z ∈ ∂M. If z ∈ Int(M) then the proof follows from the regular value theorem for manifolds without boundary, so we now focus on the case when z is a boundary point. Since M is a manifold, there exists a neighborhood U ⊂ M containing z which is diﬀeomorphic to an open subset W of the half-space Hm = m m {(x1, . . . , xm) ⊂ R | xm ≥ 0} via some diﬀeomorphism ϕ : U → W ⊂ H . Consider

19 2.1 Regular and Critical Values 2 Regular Values

the composition f˜ = f ◦ ϕ−1 : W → N. Observe that if y is a regular value of f, then the chain rule implies y is also a regular value of f˜. Set x = ϕ(z). Since f˜ is smooth, there is a neighborhood V ⊂ Rm containing x and a smooth extension h of f˜ such that ˜ ˜ ˜ h|W ∩V = f|W ∩V . Now x is a regular point of f so Dxf is surjective, which implies Dxh is also surjective since f˜ and h agree on a neighborhood of x.

M N f

z y f −1(y)

−1 Rm f ◦ ϕ W

We would like to conclude that y is a regular value for h, but this can be accomplished by m shrinking V to a smaller neighborhood of x in R . Now since Dxh is surjective, by linear algebra the matrix representing Dxh has an n × n minor with nonzero determinant. Since the determinant function is continuous (actually smooth), there is a neighborhood Ve of x 0 such that the same minor of Dx0 h has nonzero determinant for all x ∈ Ve. Therefore Dx0 h is surjective for all x0 ∈ Ve. This proves h−1(y) ∩ Ve is a manifold of dimension m − n. Finally, consider the height map

−1 g : h (y) −→ R (x1, . . . , xm) 7−→ xm.

˜−1 −1 Then f (y) = {(x1, . . . , xm) ∈ h (y) | xm ≥ 0}. To apply Lemma 2.1.8, we must show 0 is a regular value of g, but since the codomain of g is one-dimensional, it suffices to show −1 −1 that Dxg is nonzero for all x ∈ g (0). This differential is zero if and only if Txh (y) is contained in Rm−1 × {0} but as we showed, this cannot happen when y is a regular value of h. Therefore 0 is a regular value of g, and Lemma 2.1.8 tells us that f˜−1(y) is a manifold with boundary. The consequences of Theorem 2.1.9 are extremely important in the study of fixed points of smooth functions on manifolds.

20 2.1 Regular and Critical Values 2 Regular Values

Corollary 2.1.10. Let M be a compact manifold with boundary. Then there does not exist a smooth map f : M → ∂M such that f|∂M = id∂M . Proof. Suppose f : M → ∂M is smooth and fixes ∂M. By Sard’s Theorem (see Section 2.2), there is a regular value y ∈ N of f. In this case y is automatically a regular value for f|∂M since this is the identity map on ∂M, for which every value is regular. By Theorem 2.1.9, f −1(y) is a submanifold of M of dimension 1 with ∂(f −1(y)) = f −1(y) ∩ ∂M. Therefore the only point in f −1(y) ∩ ∂M is y itself, so ∂(f −1(y)) = {y}. However, f −1(y) ⊂ M and M is compact, so f −1(y) is compact. By the classification of compact 1-manifolds with boundary, f −1(y) must have two boundary points. Therefore no such f exists. This allows us to prove Brouwer’s famous fixed point theorem.

Corollary 2.1.11 (Brouwer’s Fixed Point Theorem). Any smooth map f : Dn → Dn on the closed unit n-disk Dn has a ﬁxed point.

Proof. Suppose there is a smooth map f : Dn → Dn without a fixed point. Then for every n n x ∈ D , x 6= f(x) so define Lx to be the line in R containing the distinct points x, f(x). n n ∼ n−1 Define the map g : D → ∂D = S by setting g(x) to be the intersection of Lx and ∂Dn in the closest point to x. Then g is smooth, so by Corollary 2.1.10, g cannot fix the boundary. However, for any boundary point x ∈ ∂Dn, g(x) is precisely x, a contradiction. Hence f must have a fixed point.

Example 2.1.12. Recall that the projective plane RP 2 is the quotient of S2 under the quotient map π : S2 → RP 2. The equator S1 of the sphere projects to the circle RP 1 ⊂ RP 2. Suppose there exists a smooth map f : RP 2 → R and a regular value y ∈ R such that f −1(y) = RP 1. Then y must lie in the interior of im f so there exist a, b ∈ im f with 2 1 a < y < b. Take x1, x2 ∈ RP r RP such that f(x1) = a and f(x2) = b. Note that RP 2 r RP 1 is homeomorphic to a closed disk, so it’s connected. Therefore there is a smooth 2 1 path α in RP r RP connecting x1 to x2. Then f ◦ α is a path in R connecting a to b but missing y. This of course contradicts a < y < b and the intermediate value theorem, so no such f exists.

The above example shows that the converse to the regular value theorem is false! How- ever, the converse does hold locally:

Proposition 2.1.13. If Sk is a smooth submanifold of M m and p ∈ S then there is a neighborhood U ⊂ M containing p and a smooth map f : U → Rm−k such that f −1(y) = S∩U for some regular value y of f.

Proof. By definition of a submanifold, for each point p ∈ S there is a neighborhood U ⊂ M of p and a diffeomorphism ϕ : U → ϕ(U) ⊆ Rm such that ϕ(S ∩ U) = Rk ∩ ϕ(U). Define the m m−k projection π : R → R , (x1, . . . , xk, xk+1, . . . , xm) 7→ (xk+1, . . . , xm). Then 0 is a regular value of π and π−1(0) = Rk ⊆ Rm. Then since ϕ is a diffeomorphism, ϕ−1(π−1(0) ∩ ϕ(U)) = S ∩ U.

21 2.2 Sard’s Theorem 2 Regular Values

2.2 Sard’s Theorem

The most important theorem describing regular and critical values of a smooth map is Sard’s Theorem. It guarantees the existence of regular values in general, and is the main ingredient in the theory of transversality (Section 3.1).

Deﬁnition. A set A ⊂ Rk has Lebesgue measure zero, denoted λ(A) = 0, if given any St ε > 0, there is a collection of measurable sets R1,...,Rt such that A ⊆ i=1 Ri and the sum of the measures of the Ri is less than ε.

Theorem 2.2.1 (Sard). Let f : U → Rn be a smooth function on an open set U ⊂ Rm and deﬁne the set of critical points C = {x ∈ U | rank Dxf < n}. Then f(C) has Lebesgue measure zero.

Proof. We induct on m, the dimension of the domain. For each i ∈ N, set

Ci = {p ∈ U : all partials of f up to order i are 0} ∂jf = p ∈ U : k = 0 for all j ≤ i, 1 ≤ k ≤ d . ∂x`1 ∂x`2 ··· ∂x`j

Clearly C ⊃ C1 ⊃ C2 ⊃ · · · . The proof consists of three parts:

(1) f(C r C1) has measure zero.

(2) f(Ci r Ci+1) has measure zero for all i ≥ 1.

(3) For some sufficiently large i, f(Ci) has measure zero. This will prove the theorem, since we will be able to write f(C) as a finite union of measure zero sets, and this of course implies λ(f(C)) = 0. (1) Consider a point p ∈ C r C1. Some first order partial of f is nonzero at p. Without loss of generality we may relabel the components and coordinates of f so that ∂f1 (p) 6= 0. ∂x1 Consider the map h : U → h(U) ⊂ Rn defined by

h(x1, x2, . . . , xn) = (f1(x1, . . . , xn), x2, . . . , xn).

The diﬀerential of h at p is

 ∂f1 ∗ ∗ · · · ∗ ∂x1  0 1 0 ··· 0    0 0 1 ··· 0 Dph =    ......   . . . . . 0 0 0 ··· 1

Since ∂f1 (p) 6= 0, this differential is nonsingular at p. Hence by the inverse function theorem, ∂x1 h is a diffeomorphism on a neighborhood of p, say V ⊂ U and h : V → W = h(V ) ⊂ Rn. Define g = f ◦ h−1 : W → Rd. Then since h−1 is a diffeomorphism, the critical values of g are precisely those of f, and so the critical points of g are h(C ∩ V ).

22 2.2 Sard’s Theorem 2 Regular Values

Now we claim that g maps hyperplanes to hyperplanes. To be more speciﬁc, for any t ∈ R consider ({t} × Rn−1) ∩ W , the intersection of a hyperplane in Rn with W . Then g acts on points in such a hyperplane in the following way:

−1 g(t, x2, . . . , xn) = f(h (t, x2, . . . , xn)) −1 = f(f1 (t), x2, . . . , xn) d−1 = (t, f2(~x), . . . , fd(~x)) ∈ {t} × R .

So the claim holds, and further, we can represent g as a family of functions {gt}t∈R: n−1 d−1 gt : R → R

that satisfy g(t, ~x) = (t, gt(~x)) for all t ∈ R. Take y ∈ W . Then it is easy to see that y is a critical point for g if and only if it is a critical point for each gt. By induction, since n−1 < n, the Lebesgue measure of the critical values of gt is zero but f(C ∩V ) = g(h(C ∩V )) is equal to the union [ {t} × {critical points of gt}. t∈R d Each of these sets in the union is a subset of a hyperplane intersecting f(C r C1) in R , each of which has measure zero, so this implies by properties of Lebesgue measure that f(C r C1) has measure zero. (2) The proof that f(Ci rCi+1) has measure zero is essentially the same as (1). This time, we may assume that ∂ ∂if 6= 0 ∂x1 ∂x`1 ··· ∂xì ∂if and define h(x1, . . . , xn) = (x1, . . . , xn), x2, . . . , xn . ∂x`1 ··· ∂xì (3) Take x ∈ Ci. The idea is to take a small n-cube around x, subdivide it into smaller cubes and compare the volumes of the cube and its subdivisions. Let K be an n-cube around x with side length δ:

K x δ

Take h ∈ Rn such that x + h ∈ K. By Taylor’s Theorem, since the ﬁrst i partials of f at x are zero, we can write

i+1 i+1 i+1 f(x + h) = f(x) + R(x, h), where |R(x, h)| < C||h|| < Cδ n 2 . Further subdivide the sides of K into r equal subintervals to give rn subcubes, each with side δ length r . Consider the subcube K1 ⊂ K containing x. By a similar estimate using Taylor’s Theorem, for any h such that x + h ∈ K, i+1 i+1 δ i+1 1 f(x + h) = f(x) + R(x, h) where |R(x, h)| < C||h|| < C n 2 = κ , ri+1 ri+1

23 2.3 Homotopy and Degree mod 2 2 Regular Values

for a constant κ which depends on n and δ. This calculation shows that the image under f d κ of the cube K1 is contained in a d-cube in R around f(x) with side length less than ri+1 . Consider f(Ci ∩ K1). This has measure κ d λ(f(C ∩ K )) < rn = κdrn−d(i+1) i 1 ri+1 which must approach 0 as r → ∞. By the above formula for λ(f(Ci ∩ K1)), this can only n happen if n−d(i+1) < 0, or i > d −1. So there is a suﬃciently large i satisfying the desired property. This concludes the proof of Sard’s Theorem.

2.3 Homotopy and Degree mod 2

To prove the well-deﬁnedness of deg2 f, we will actually prove a more general statement: Theorem 2.3.1. If f, g : M n → N n are smoothly homotopic maps between compact, smooth

manifolds without boundary, then deg2 f = deg2 g. This requires the notion of smooth homotopy. The definition is similar to the definition of homotopy in algebraic topology, but with the additional condition that the homotopy map is smooth. Rigorously: Definition. Suppose X and Y are smooth manifolds and f, g : X → Y are smooth maps. Then f and g are smoothly homotopic, denoted f ' g, if there exists a smooth map F : X × [0, 1] → Y such that F (x, 0) = f(x) and F (x, 1) = g(x) for all x ∈ X. Such a map F is called a smooth homotopy.

X g Y

[0, 1] t ft

X f

F may be viewed as a one-parameter family of maps F (x, t) = ft(x): X → Y such that each ft(x) is smooth. Proposition 2.3.2. Smooth homotopy is an equivalence relation. When considering maps X → X, there is a stronger notion of equivalence known as smooth isotopy. This requires that the family of maps ft are all diffeomorphisms. Definition. Two maps f, g : X → X are smoothly isotopic if there exists a smooth homotopy F : X ×[0, 1] → X, i.e. a one-parameter family of maps F (x, t) = ft(x): X → X, such that f0 = f, f1 = g and for all t ∈ [0, 1], ft is a diffeomorphism on X. Such a map F is called a smooth isotopy.

24 2.3 Homotopy and Degree mod 2 2 Regular Values

Example 2.3.3. When X = S1, there are only two diffeomorphisms from S1 to itself: the identity id : eit 7→ eit and the antipodal map a : eit 7→ e−it. In this case a and id are not smoothly homotopic (and hence not isotopic). Example 2.3.4. The Dehn twist is a diffeomorphism on the torus that is not isotopic to 2 the identity. (The figure below depicts a half-twist, so the diffeomorphism is fD.)

As a first step towards proving that degree mod 2 is well-defined, we will now prove the Homotopy Lemma. Lemma 2.3.5 (Homotopy Lemma). Suppose f, g : M → N are smoothly homotopic maps between compact manifolds without boundary of the same dimension. If y ∈ N is a regular value for both f and g, then the cardinalities of f −1(y) and g−1(y) are congruent mod 2. Proof. Suppose f and g are smoothly homotopic. Then there exists a smooth homotopy −1 F : M × [0, 1] → N such that F |M×{0} = f and F |M×{1} = g. Since M is compact, |f (y)| and |g−1(y)| are each finite. Assume y is also a regular value for F (we will justify this in a moment). Then by the regular value theorem, F −1(y) is a compact, one-dimensional submanifold of M × [0, 1] with boundary. Specifically:

∂F −1(y) = F −1(y) ∩ ∂(M × [0, 1]) = F −1(y) ∩ (M × {0} ∪ M × {1}) = f −1(y) ∪ g−1(y).

The only compact 1-manifolds with boundary disjoint unions of circles and closed intervals, so in particular, the boundary of F −1(y) consists of an even number of points. Either a boundary component can connect two points in f −1(y) (or two points in g−1(y)) or it can connect one point in f −1(y) with a point in g−1(y). This implies |f −1(y)| ≡ |g−1(y)| mod 2. We now address the concern that y may not be a regular value for F . Recall from Lemma 2.1.1 that |f −1(y)| is a locally constant function of y. Then there is a neighborhood U of y such that for all y0 ∈ U, |f −1(y0)| = |f −1(y)|. For the map g, there is a similar neighborhood V of y such that for all y0 ∈ V , |g−1(y0)| = |g−1(y)|. The intersection U ∩ V is an open neighborhood of y on which |f −1(y0)| and |g−1(y0)| are each constant. By Sard’s Theorem (2.2.1), the set of regular values of F has full measure in N, so U ∩ V contains regular values of F . Now we can run the proof above for such a regular value y0 ∈ U ∩V ⊂ N and obtain the result that |f −1(y0)| ≡ |g−1(y0)| mod 2 which implies the result. Next we prove the powerful Homogeneity Lemma, which has applications reaching far beyond the study of degree mod 2. Lemma 2.3.6 (Homogeneity Lemma). Suppose N is a connected manifold, possibly with boundary, and let y, z ∈ N r ∂N. Then there exists a diﬀeomorphism f : N → N such that f(y) = z and f is smoothly isotopic to the identity on N.

25 2.3 Homotopy and Degree mod 2 2 Regular Values

Proof. We first prove the lemma locally, i.e. on a neighborhood U of y. Then in the general case, we will ‘bootstrap’ together neighborhoods to get from y to z. Consider a neighborhood U ⊂ N containing y. Then there is a diffeomorphism ϕ : U → ϕ(U) ⊂ Rn. We can rescale and translate ϕ so that ϕ(U) is the open unit ball Bn centered at y = 0, so we will prove the statement on Bn and transport everything back to the manifold via ϕ−1. There exists a smooth function g : (0, ∞) → R (a certain type of exponential function) such that g ≡ 0 on [1, ∞) and g is strictly positive on (0, 1). Use g to construct a map ψ : Rn → R such that ψ(¯x) > 0 forx ¯ ∈ Bn and ψ(¯x) = 0 forx ¯ 6∈ Bn – explicitly, 2 2 n the construction is ψ(x1, . . . , xn) = g(x1 + ... + xn). Given some z ∈ B , we use ψ to define a diffeomorphism f : Rn → Rn which is isotopic to the identity and satisfies f(0) = z. n Consider the unit vector v in R with direction z; write v = (v1, . . . , vn). Then we have a system of ODEs  dx1 = ψ(x , . . . , x )v  dt 1 n 1  dx2  dt = ψ(x1, . . . , xn)v2 (∗) = .  .   dxn dt = ψ(x1, . . . , xn)vn. (The system can also be writtenx ¯0(t) = ψ(¯x)v.) The theory of ODEs says that there exists n a unique smooth solutionx ¯(t) with the initial conditionx ¯(0) = x0 for any x0 ∈ R . Further, since these solutions depend smoothly on the initial condition, we have a functionx ¯(t, x0) which is smooth as a function of t and x0. This function is a good candidate for the smooth isotopy we would like to define. For a fixed value of t,x ¯(t, x0) andx ¯(−t, x0) are smooth inverses of each other, so they are both diffeomorphisms (¯x(−t, x0) is the unique solution to the system of ODEs obtained by multiplying (∗) through by −1). Now taking x0 = 0, there is some t0 ∈ R such thatx ¯(t0, 0) = z by construction. Then fixing t0 but letting x0 n n n range over all points in R , we obtain a diffeomorphismx ¯(t0, x0): R → R that is isotopic to the identity,x ¯(0, 0). We have therefore proven the lemma on Bn and we can translate everything back up to U ⊂ N using ϕ−1. For the global case, define an equivalence relation on N by y ∼ z if there exists a diffeomorphism N → N which is smoothly isotopic to idN and takes y 7→ z. By the work above, each equivalence class under ∼ is an open set and the collection of these partitions N, but since N is connected, there can only be a single equivalence class. Therefore y ∼ z for all y, z ∈ N, proving the lemma. Now we can prove the well-definedness of degree mod 2.

Theorem 2.3.7. If y, z ∈ N are two regular values of f : M m → N n then |f −1(y)| and |f −1(z)| are congruent mod 2.

Proof. By the Homogeneity Lemma, there exists a diffeomorphism ϕ : N → N which is isotopic to idN and satisfies ϕ(y) = z. Set g = ϕ ◦ f. Then f and g are smoothly homotopic via the homotopy F = ϕ◦(f ×id[0,1]): M ×[0, 1] → N. Moreover, since ϕ is a diffeomorphism and y is a regular value of f, y is also a regular value of g = ϕ ◦ f. By the Homotopy Lemma, |f −1(z)| ≡ |g−1(z)| mod 2 but by construction, g−1(z) = f −1(y). This proves the theorem.

26 2.4 Orientation 2 Regular Values

2.4 Orientation

Returning to our motivating example in this chapter, consider S1 ⊂ C, the unit circle sitting 1 1 iθ niθ in the complex plane. Define a family of functions fn : S → S by fn(e ) = e . Intuitively, the winding number of fn ought to be n. We want this to correspond to a general notion of degree, one which takes values over the integers. To define such a degree, we first need to introduce a rigorous notion of orientation. Example 2.4.1. As another motivating example, take a two-component link L : S1 t S1 ,→ R3. The linking number `k(L) is a topological invariant which describes the (signed) number of times the two components cross each other in a two-dimensional diagram of the link. If f : S1 ,→ R3 and g : S1 ,→ R3 are the two smooth embeddings of the components of L, we can define a smooth map

F : S1 × S1 −→ S2 f(x) − g(y) (x, y) 7−→ . ||f(x) − g(y)|| It turns out that the degree of F is precisely the linking number `k(L). Obviously this construction depends on which way the ‘arrows’ are chosen on the components of L. This motivates the study of orientation on a manifold. Before deﬁning orientation on manifolds, we introduce the notion of an orientation on a vector space.

Deﬁnition. An orientation on Rn is an equivalence relation of ordered bases, where B and C are equivalent ordered bases if the change of basis matrix [T ]B→C has positive determinant.

Example 2.4.2. We illustrate in R2. Fix the standard basis on R2, E = {(1, 0), (0, 1)}. Which of the following ordered bases have the same orientation as E?:

B1 = {(0, 1), (1, 0)}, B2 = {(−1, 0), (0, 1)}, B3 = {(0, 1), (−1, 0)}. Consider the determinant of each change of basis matrix:

0 1 det[T ]E→B1 = = −1; 1 0

−1 0 det[T ]E→B2 = = −1; 0 1

0 −1 det[T ]E→B3 = = 1. 1 0

This shows that B1 and B2 have the opposite orientation as E, but B3 has the same orientation as the standard basis. In general, there are two orientations on Rn:

{B | det[T ]E→B > 0} and {B | det[T ]E→B < 0}. We will call these the positive orientation and negative orientation, respectively.

27 2.5 Brouwer Degree 2 Regular Values

Definition. An oriented manifold is a smooth manifold M equipped with an orientation on TxM for each x ∈ M such that there is a neighborhood U ⊂ M of x and a diffeomorphism n m ϕ : U → ϕ(U) ⊂ R with the property that dpϕ : TpM → R takes the orientation on TpM to the same orientation on Rm for every p ∈ U. Example 2.4.3. The Möbiusband is a quotient of a rectangle:

The projective plane RP 2 naturally can be formed by gluing the M¨obiusband to a disk. Alternatively, one may glue together the remaining unidentiﬁed sides of the rectangle above to obtain the projective plane:

RP 2

Once we show that the M¨obiusband is non-orientable, this will imply that RP 2 is non- orientable as well. Observe the following circle sitting inside the band:

If B were orientable, then for every point x on this circle, we could choose an ordered basis 2 {v1, v2} of TxB so that dxϕ : TxB → R takes the orientation of {v1, v2} to the orientation of the standard basis {e1, e2}. However, at the point of gluing, e2 ﬂips to −e2 so orientation is not preserved globally on B.

This proves:

Proposition 2.4.4. The M¨obiusband and the projective plane are non-orientable manifolds.

2.5 Brouwer Degree

Deﬁnition. Suppose f : M n → N n is a smooth map between compact, oriented manifolds M,N of the same dimension. Then the Brouwer degree of f is X deg f = sign(Dxf) x∈f −1(y)

where y ∈ N is a regular value of f and sign(Dxf) = +1 if Dxf : TxM → TyN is orientation- preserving and sign(Dxf) = −1 if Dxf is orientation-reversing.

28 2.5 Brouwer Degree 2 Regular Values

Remark. As with the definition of degree mod 2, Sard’s Theorem (2.2.1) guarantees the existence of such a regular value y ∈ f(M). Moreover, the definition of sign(Dxf) requires a choice of orientations on TxM and TyN, but we will see how sign(Dxf) changes with each choice of orientation. Proposition 2.5.1. For a smooth map f : M n → N n between compact, oriented manifolds, the degree mod 2 of f is precisely the Brouwer degree of f taken mod 2. P Proof. We can write deg2 f = x∈f −1(y)(+1) (mod 2), and switching any of these +1’s to a −1 does not change the parity of the sum. We will generalize the two main properties of degree mod 2 from Section 2.3: (1) deg f is well-defined in terms of regular values y ∈ N. (2) If f, g : M → N are smoothly homotopic, then deg f = deg g. Remark. If M n is an oriented manifold with boundary then ∂M is an oriented manifold whose orientation is induced by the orientation on M. Formally, at each point x ∈ ∂M the tangent space TxM is an n-dimensional vector space consisting of outward pointing vectors, inward pointing vectors and a subspace Tx(∂M) of vectors tangent to the boundary itself. n Under a local diffeomorphism ϕ at x, these look like the following picture in R+:

ϕ(U)

inward tangent Rn−1

outward ϕ(Tx(∂M))

If B = {v2, . . . , vn} is a basis of Tx(∂M), we say that B has positive orientation (or orientation consistent with the orientation of M) if for any outward pointing vector v1 ∈ TxM, the basis {v1, v2, . . . , vn} of TxM has positive orientation, or the orientation consistent with the standard basis on Rn. Otherwise B is said to have negative orientation. Example 2.5.2. For M = D2, the closed unit disk in R2, the boundary is ∂M = S1, the unit circle. If we give D2 the ‘standard’ orientation as a subset of R2, the orientation induced on S1 is the counterclockwise orientation:

29 2.5 Brouwer Degree 2 Regular Values

Lemma 2.5.3. If M,N are orientable then M × N is orientable.

Proof. By Corollary 1.3.4, T(x,y)(M × N) = TxM ⊕ TyN. Corollary 2.5.4. If M is an oriented manifold without boundary, then M×[0, 1] is orientable and, given any orientation on M×[0, 1], the orientations induced on the boundary components M × {0} and M × {1} are opposites.

Proof. The ﬁrst statement is a direct consequence of Lemma 2.5.3. Choose an orientation on M. This induces an orientation on M0 = M ×{0} and M1 = M ×{1}. For (x, t) ∈ M ×[0, 1], ∼ Corollary 1.3.4 tells us that T(x,t)(M ×[0, 1]) = TxM ⊕R which has some basis {v1, . . . , vm, 1} where m = dim M. If {v2, . . . , vm, vm+1} is a basis for T(x,0)M0 then vm+1 = c1 for some scalar c ∈ R so we may assume vm+1 = 1. Then given such a basis {v2, . . . , vm, 1} of T(x,0)M0, any outward pointing vector v1 ∈ T(x,0)(M × [0, 1]) is an inward pointing vector in T(x,1)(M×[0, 1]). Thus {v2, . . . , vm, 1} has opposite orientations for T(x,0)M0 and T(x,1)M1. Lemma 2.5.5. Suppose M and N are oriented manifolds and f, g : M → N are smoothly homotopic maps with smooth homotopy F : M × [0, 1] → N, and suppose y ∈ N is a regular value for f, g and F . Then X X sign(Dxf) = sign(Dxg). x∈f −1(y) x∈g−1(y)

Proof. The proof is nearly the same as in the degree mod 2 case. As in the proof of the homotopy lemma , we may assume y = 0 ∈ R. Then F −1(0) ⊂ M × [0, 1] is a compact one- −1 dimensional manifold with an induced orientation. Let v be a nonzero vector in TxF (0) for −1 −1 some x ∈ F (0). This tangent space is a one-dimensional vector space so v spans TxF (0). By the remark, v defines the induced orientation on F −1(0) if and only if when we extend {v} to a basis {v, v2, . . . , vm+1} of M ×[0, 1] it has the orientation defined on M ×[0, 1]. Now since DxF is a linear isomorphism, this is equivalent to {DxF (v2),...,DxF (vm+1)} having the chosen orientation on N. Take A to be an arc in F −1(0), i.e. one of the components of this compact 1-manifold with boundary. The construction above defines a smooth, nonvanishing vector field {v(t)} on F −1(0) where each v(t) is consistent with the ambient orientation of M × [0, 1]. Consider a parametrization α : [0, 1] → A with α0(t) 6= 0 for all t ∈ [0, 1]. We claim α0(0) and α0(1) define the opposite orientation on A. Notice that

α(t) − α(0) α0(0) = lim t→0 t and α(t) is on the interior of M × [0, 1], so α0(0) must be inward pointing. For the same reason, α0(1) must be outward pointing. Now it is clear that α0(t) is parallel to v(t) for each t, and this is consistent with the orientation on the arc A. Thus the two endpoints of A have opposite orientations. This implies that the signs of the diﬀerentials at the two endpoints are opposite. To see this more explicitly, consider the case where the endpoints x0, x1 of A both lie in M0 (the other cases are analagous). Then v0 = v(x0) points inward and v1 = v(x1) points outward. Complete v0 to a positively-oriented basis {v0, v2, . . . , vm+1}

30 2.5 Brouwer Degree 2 Regular Values

of Tx0 (M × [0, 1]). Then since v0 points inward, {v2, . . . , vm+1} is a negatively-oriented basis of Tx0 (M0). The same argument at x1 shows that the same basis {v2, . . . , vm+1} is positively-

oriented for Tx1 (M1). Therefore sign(Dx0 F ) = − sign(Dx1 F ). Further, since F |∂(M×[0,1]) = f, this implies that

sign(Dx0 f) = − sign(Dx1 f). −1 Now consider the collection of arcs in F (0). For each arc with both endpoints in M0 (or M1), the sum of the signs of Dxf (or Dxg) is not changed since the endpoints have opposite signs. For an arc with an endpoint in both M0 and M1, a +1 is contributed to one sum and a −1 to the other sum, but by Corollary 2.5.4, M0 and M1 hafve opposite orientations so these really contribute the same value to each sum. Therefore X X sign(Dxf) = sign(Dxg). x∈f −1(0) x∈g−1(0)

Theorem 2.5.6. Brouwer degree is well-defined: if y, z ∈ N are two regular values of P P f : M → N then x∈f −1(y) sign(Dxf) = x∈f −1(z) sign(Dxf). Proof. The homogeneity lemma is the same in the oriented case, so we can again find a diffeomorphism ϕ : N → N isotopic to idN taking y 7→ z. Then f is smoothly homotopic to ϕ ◦ f and y is a regular value for ϕ ◦ f, so by Lemma 2.5.5, X X X sign(Dxf) = sign(Dx(ϕ ◦ f) = sign(Dxf). x∈f −1(z) x∈(ϕ◦f)−1(z) x∈f −1(y)

Hence Brouwer degree is well-deﬁned. Examples. 1 For every n ≥ 1, Sn is orientable. If we view Sn as the preimage of 1 under the function 2 2 ∇f(~x) n f(x1, . . . , xn+1) = x1 + ... + xn+1, then ||∇f(~x)|| is a unit normal vector to S at each ~x ∈ Sn. It’s easy to see that ∇f(~x) = ~x for each ~x ∈ Sn. Moreover, Sn = ∂Bn+1 where Bn+1 is the closed unit (n + 1)-ball. As Bn+1 ⊂ Rn+1, Bn+1 has a natural orientation so Sn inherits an orientation from the standard orientation on Bn+1. 1 1 iθ niθ Consider the family of functions fn : S → S deﬁned for each n ∈ Z by fn(e ) = e . Then deg fn = n.

n n 2 Define the reflection map r1 : S → S by r1(x1, x2, . . . , xn+1) = (−x1, x2, . . . , xn+1) and similarly define ri by reflection in the ith component for 1 ≤ i ≤ n + 1. Each of these maps has degree −1, while the identity map always has degree 1. The antipodal n n map a : S → S , ~x 7→ −~x, may be viewed as the composition a = r1 ◦ r2 ◦ · · · ◦ rn+1. Then by the multiplicative property of Brouwer degree, deg a = (−1)n+1.

3 We have maps from S2 to itself of degrees 0 (constant map), 1 (identity) and −1 (antipodal map). How do we construct a map of arbitrary degree? The following construction is called a suspension of a topological space X. Deﬁne ΣX = X ×[0, 1]/ ∼

31 2.5 Brouwer Degree 2 Regular Values

where ∼ identiﬁes all points in X × {0} to a single point and all points in X × {1} to a diﬀerent point. For example, ΣX1 is two cones glued together in a shape that is homeomorphic to S2:

S1 × {1}

ΣS1 ∼= S2

S1 × {0}

Any smooth map f : X → Y between manifolds induces a smooth map on their suspensions:

Σf :ΣX −→ ΣY (x, t) 7−→ (f(x), t).

1 1 n 1 For fn : S → S , z 7→ z , we can view ΣS diffeomorphically as the two-sphere by smoothing out the two cones when the points are identified under ∼. Then Σfn : ΣS1 → ΣS1 gives us a map of degree n on S2, for any integer n. Taking further suspensions and smoothing the resulting figures at the poles allows us to construct maps of arbitrary degree from Sk to itself for any k ≥ 1.

4 Given disjoint manifolds M m,N n in Rk+1, the linking map λ : M × N → Sk is x − y λ(x, y) = . |x − y| If M,N are compact, oriented, and without boundary, and m + n = k, then the Brouwer degree of λ is called the linking number l(M,N). We will show that l(N,M) = (−1)(m+1)(n+1)l(M,N). Set y − x γ : N × M → Sk, γ(y, x) = . |y − x|

Then we can interpret the linking numbers in the desired formula as l(M,N) = deg λ and l(N,M) = deg γ. Consider the map f : N × M → M × N, (y, x) 7→ (x, y). Then clearly γ = (−λ) ◦ f so by the multiplicative property of Brouwer degree,

32 2.5 Brouwer Degree 2 Regular Values

deg γ = deg(−λ) deg f. Starting with f, we see that on the level of tangent spaces, the diﬀerential of f acts by swapping the M and N parts of the tangent space:

d(y,x)f : TyN ⊕ TxM −→ TxM ⊕ TyN

(v1, . . . , vn, w1, . . . , wm) 7−→ (w1, . . . , wm, v1, . . . , vn).

This changes the sign of the diﬀerential by −1 for each pair of components that is mn swapped, i.e. the sign changes by (−1) . Next, by linearity dx(−λ) = −dxλ so the k+1 sign of this diﬀerential is (−1) times the sign of dxλ since sign is calculated using a (k + 1) × (k + 1) determinant in Rk+1. Finally, putting these together and using the assumption that m + n = k gives us

l(N,M) = deg γ = (−1)mn(−1)k+1 deg λ = (−1)mn+m+n+1 deg λ = (−1)(m+1)(n+1)l(M,N).

33 3 Transversality and Embeddings

3 Transversality and Embeddings

3.1 Transversality, Immersion and Submersion

The notion of transversality, or “general position”, is a fundamental topic in diﬀerential topology. It generalizes the idea of regularity and relies on Sard’s Theorem (2.2.1) extensively. Deﬁnition. Suppose M is a smooth manifold of dimension m and S and T are smooth submanifolds of M of dimensions s and t, respectively. We say S and T are transverse (or in general position), denoted S t T , if for every p ∈ S ∩ T , the vector space spanned by TpS and TpT is TpM.

Example 3.1.1. Here are some examples of one-dimensional submanifolds in R2: T T T

S S S

transverse transverse not transverse

Example 3.1.2. In R3, two 2-submanifolds can intersect in general position in various ways, but two 1-submanifolds are in general position if and only if they are disjoint. This is easy to see from the definition: if S and T are 1-manifolds then TxS and TxT are 1-dimensional vector spaces for any x ∈ S ∩ T so they cannot generate the 3-dimensional vector space 3 3 TxR = R . The main theorem in the theory of transversality is the following. Theorem 3.1.3. Suppose Ss,T t ⊂ M m are all compact, smooth manifolds embedded in Euclidean space. Then there exists a smooth ε-isotopy ϕ : M → M, i.e. an isotopy that moves all points by a distance less than ε, such that ϕ(T ) is transverse to S. There are two special classes of smooth maps between manifolds which are of particular interest in differential topology. Definition. A smooth map f : M m → N n is an immersion at a point p ∈ M if the differential Dpf : TpM → Tf(p)N is injective. We say f is an immersion if it is an immersion at every point p ∈ M. Clearly it is only possible to define an immersion if m ≤ n. Proposition 3.1.4. If f : M → N is an immersion at a point p ∈ M then there exists a neighborhood U ⊂ M of p such that f is an immersion at every x ∈ U.

34 3.1 Transversality, Immersion and Submersion 3 Transversality and Embeddings

Proof. The diﬀerential Dpf is injective if and only if in the n×m matrix A representing Dpf, there is an m × m minor with nonzero determinant. Since the determinant is a continuous function (in fact, it is smooth), this same minor has nonzero determinant on a neighborhood of p. Hence Dxf will be injective on this neighborhood. Examples.

1 Suppose M ⊂ N is a submanifold. Then by deﬁnition for every x ∈ M, there is a neighborhood U ⊂ N of x and a diﬀeomorphism ϕ : U → ϕ(U) ⊆ Rn such that ϕ(U ∩ N) ⊆ Rm × {0¯} ⊆ Rm × Rn−m. Consider the embedding i : M,→ N. For any x ∈ M, Dxi : TxM,→ TxN is an injection, so i is an immersion. For example, if m n m ≤ n then the canonical inclusion R ,→ R , (x1, . . . , xm) 7→ (x1, . . . , xm, 0,..., 0) is an immersion.

2 There are immersions that are not embeddings. For example, the degree n maps 1 1 n fn : S → S , z 7→ z are immersions (the tangent line to the circle is nondegenerate everywhere) but when |n| ≥ 2, fn is not one-to-one.

3 A parametrized diﬀerential curve is the image of a map g : [0, 1] → Rn such that g0(t) 6= 0 for all t. The map g is an immersion and in many cases may not be an embedding.

We next deﬁne the related notion of a submersion.

Deﬁnition. A smooth map f : M m → N n is a submersion at a point p ∈ M if the diﬀerential Dpf : TpM → Tf(p)(N) is surjective. We say f is an submersion if it is an submersion at every point p ∈ M.

Clearly it is only possible to deﬁne a submersion if m ≥ n.

Proposition 3.1.5. If f : M → N is a submersion at a point p ∈ M then there exists a neighborhood U ⊂ M of p such that f is a submersion at every x ∈ U.

Proof. Identical to the proof for immersions. Examples.

n+m n 4 The coordinate projections R → R , (x1, . . . , xn, xn+1, . . . , xn+m) 7→ (x1, . . . , xn) for all n, m ≥ 0 are canonical examples of submersions. More generally, for any smooth manifolds M,N, the coordinate projections M × N → M and M × N → N are submersions.

m 5 Suppose that f1, . . . , fk : M → R are smooth functions on a smooth manifold M and deﬁne their vanishing locus

Z = {p ∈ M | fi(p) = 0 for all 1 ≤ i ≤ k}.

k ¯ Deﬁne the map f = (f1, . . . , fk): M → R . Then Z is the preimage of the point 0 = k (0,..., 0) ∈ R . Since each fi is smooth, Dpfi : TpM → R is linear. Moreover, one can

35 3.1 Transversality, Immersion and Submersion 3 Transversality and Embeddings

k see that Dpf = (Dpf1,...,Dpfk): TpM → R is surjective if and only if Dpf1,...,Dpfk k are linearly independent on TpM. Therefore f : M → R is a submersion if and only if the Dpfi are linearly independent at every p ∈ M, and in this case, the regular value theorem (2.1.2) implies Z is a smooth submanifold of M of dimension m − k. When the fi are all polynomials, Z is called an algebraic variety. The next theorem is fundamental in the study of immersions and submersions. Intuitively, it says that every immersion is locally a coordinate inclusion (as in Example 1 ) and every submersion is locally a coordinate projection (as in Example 4 ).

Theorem 3.1.6. Let f : M → N be a smooth map and take p ∈ M.

(1) If f is an immersion at p then there exist coordinate charts (U, ϕ) around p and (V, ψ) around f(p) such that in these local coordinates, the map ψ◦f ◦ϕ−1 : ϕ(U) ⊂ Rm → Rn is of the form x¯ 7→ (¯x, 0)¯ ∈ Rm × Rn−m. (2) If f is a submersion at p then there exist coordinate charts (U, ϕ) around p and (V, ψ) around f(p) such that in these local coordinates, ψ ◦ f ◦ ϕ−1 is of the form (x1, . . . , xn, xn+1, . . . , xm) 7→ (x1, . . . , xn). Proof. We prove (1) and remark that the proof of (2) is similar. Consider any charts (U, ϕ) and (V, ψ) around p and f(p), respectively. Then at ϕ(p) ∈ ϕ(U) ⊆ Rm, the differential of −1 h = ψ ◦ f ◦ ϕ is injective by the chain rule. In particular, im Dϕ(p)h is an m-dimensional subspace of Rn. We may compose with an isometry L (a rotation) of Rn to arrange it so m n that im Dϕ(p)h lies in the subspace R × {0} ⊆ R . Replacing (V, ψ) with (V,L ◦ ψ), we may just assume that this situation occurs to begin with. m Now the differential Dϕ(p)h takes the tangent space at ϕ(p) into R × {0}. We next change the coordinates so that the entire image of U lies in Rm × {0}. First define a map A : Rn → Rn so that for allx ¯ ∈ Rm, A(¯x, 0)¯ = h(¯x). Extend this to all of Rn = Rm × Rn−m by A(¯x, y¯) = h(¯x)+y ¯. Then A is a diffeomorphism so the charts (U, ϕ) and (A−1(V ),A−1 ◦ψ) satisfy the conclusion of the theorem. If f is both an immersion and a submersion at p ∈ M then of course p is a regular point of f. The following theorem generalizes the regular value theorem (Theorem 2.1.2).

n1 n2 m Theorem 3.1.7. Suppose N1 and N2 are submanifolds of a smooth manifold M such that N1 t N2. Then N1 ∩ N2 is a submanifold of M of dimension n1 + n2 − m. Moreover, for every p ∈ N1 ∩ N2 there exist coordinate systems on N1 and N2 such that N1 corresponds n1 n2 to R × {0} and N2 corresponds to {0} × R .

Proof. Fix p ∈ N1 ∩ N2. Since N1 is a submanifold, there exists a neighborhood U1 ⊂ M m n1 of p and a diﬀeomorphism ϕ1 : U1 → ϕ1(U1) ⊂ R such that ϕ1(U1 ∩ N1) ⊆ R × {0}. m m−n1 m−n1 Deﬁne the orthogonal projection π1 : R → R and set ψ1 : π1 ◦ ϕ1 : U1 → R . −1 Then 0 is a regular value of ψ1 so by the regular value theorem, N1 ∩ U1 = ψ1 (0) is a submanifold of M of dimension n1. Performing the same construction with N2 gives us a

36 3.1 Transversality, Immersion and Submersion 3 Transversality and Embeddings

m−n2 −1 map ψ2 : π2 ◦ ϕ2 : U2 → R such that U2 ∩ N2 = ψ2 (0) is a submanifold of M of dimension n2. Set U = U1 ∩ U2 so that ψ1 and ψ2 are both deﬁned on U. Deﬁne a map

m−n m−n 2m−n −n ψ : U −→ R 1 × R 2 = R 1 2 x 7−→ (ψ1(x), ψ2(x)).

−1 Now N1 ∩ N2 ∩ U = ψ (0, 0) so it suﬃces to show that (0, 0) is a regular value of ψ, for then N1 ∩ N2 will be a submanifold of M of dimension m − (2m − n1 − n2) = n1 + n2 − m. m−n1 m−n2 −1 Consider the diﬀerential Dpψ : TpM → R × R for any p ∈ ψ (0, 0). Then m−n1 Dpψ = (Dpψ1,Dpψ2) componentwise and by the above, Dpψ1 is surjective onto R and m−n2 Dpψ2 is surjective onto R . By Corollary 2.1.6, ker(Dpψ) = ker(Dpψ1) ∩ ker(Dpψ2) = TpN1∩TpN2, but by transversality, TpN1 and TpN2 span TpM. So dim ker(Dpψ) ≥ n1+n2−m. However, it is always true that dim ker(Dpψ) ≤ n1 + n2 − m, so we have dim ker(Dpψ) = n1 + n2 − m. Hence by rank-nullity,

dim im(Dpψ) = m − (n1 + n2 − m) = 2m − n1 − n2, so Dpψ is surjective. This shows that (0, 0) is a regular value of ψ so we conclude that N1 ∩ N2 is a submanifold of the desired dimension. For the second statement, use the fact that N1 ∩ N2 is a submanifold to choose a neighborhood V ⊂ M containing p and a diﬀeomorphism φ : V → φ(V ) ⊆ Rm such that n1+n2−m n1+n2−m φ(V ∩ N1 ∩ N2) ⊆ R × {0}. Extend the target to R by projection and abuse notation by letting f denote this function V → Rn1+n2−m. Consider the map

m−n n +n −m m−n m f : U −→ R 2 × R 1 2 × R 1 = R x 7−→ (ψ2(x), φ(x), ψ1(x)).

We claim that f is a diffeomorphism satisfying the conditions of the theorem. First, consider m−n2 n1+n2−m m−n1 the differential Dpf : TpM → R × R × R . As subspaces of TpM, TpN1 ∩ n1+n2−m TpN2 = Tp(N1 ∩ N2). If v ∈ ker(Dpψ2) ∩ ker(Dpψ1) = Tp(N1 ∩ N2) then φ(v) ∈ R , but ψ is a diffeomorphism so in this case v = 0. This implies that Dpf is nonsingular, and −1 −1 thus an isomorphism. Finally, since N1 ∩ U = ψ1 (0) and N2 ∩ U = ψ2 (0), it follows that n1 n2 f(N1 ∩ U) ⊆ R × {0} and f(N2 ∩ U) ⊆ {0} × R .

Example 3.1.8. The torus T = S1 × S1 does not admit an immersion into R2, but the punctured torus T ∗ does have this property. Consider the torus as the quotient space of a square, and expand the removed disk so that it almost captures all of the interior of the square:

37 3.2 Embedding Theorems 3 Transversality and Embeddings

The resulting quotient space T ∗ = T r D looks like two interlocking bands joined together at one small square. Alternatively, these bands can be viewed as thickened strips around a meridian and longitude of the torus. To obtain an immersion f : T ∗ → R2, simply embed T ∗ into R3 in a way such that there are no vertical tangent planes to T ∗ – note that this is not possible with the unpunctured torus, but removing a disk makes it possible. Then deﬁne f : T ∗ → R2 to be the restriction of the projection R3 → R2, (x, y, z) 7→ (x, y) to ∗ ∗ ∗ T . By construction, for every point p ∈ T the tangent plane TpT is spanned by two vectors v1 = (x1, y1, z1) and v2 = (x2, y2, z2) with at least one of x1, y1 nonzero and at least one of x2, y2 nonzero. The diﬀerential dpf takes these vectors to dp(v1) = (x1, y1) and 2 2 ∗ dp(v2) = (x2, y2) in Tf(x)R = R , and since v1 and v2 were assumed to span TpT ,(x1, y1) and 2 (x2, y2) cannot be parallel. Therefore dp(v1) and dp(v2) span R , so dpf is an isomorphism and in particular injective. Hence f is an immersion.

3.2 Embedding Theorems

Our initial definition of a smooth manifold M was as a subset of some Euclidean space Rk, where k ≥ dim M, but we later generalized this in Section 1.3 to allow more abstract manifolds such as projective n-space RP n. One may wonder if every abstract smooth manifold may be viewed as a subset of some large enough Euclidean space. Whitney’s embedding theorem gives a positive answer to this, and even more, gives a bound on the dimension k required for such a realization to exist. Thus our original definition of a manifold was completely justified. For a smooth n-manifold M, there is a particular manifold TM of dimension 2n defined using the tangent structure on M:

TM = {(x, v) | x ∈ M, v ∈ TxM}.

Then TM, called the tangent bundle to M, is smooth and has a natural map π : TM → −1 M, (x, v) 7→ x such that each preimage π (x) is precisely the tangent space TxM. We will study TM further in Chapter 4. The following theorem is a preliminary version of Whitney’s theorem, but it already answers the question of whether a manifold has an embedding into Euclidean space.

Theorem 3.2.1. Every manifold of dimension n admits an injective immersion into R2n+1.

Proof. Suppose M ⊆ RN is a smooth manifold of dimension n. We show by induction that for all M > 2n + 1, if f : M → RM is an injective immersion then there exists a unit vector M f M projv ⊥ v ∈ R such that M −→ R −−−→ Span{v} is still an injective immersion, where projv ⊥ ∼ M−1 denotes orthogonal projection onto the orthogonal complement of v. As Span{v} = R , this proves by induction that M admits an injective immersion M,→ R2n+1. Given f : M → RM as above, deﬁne h : X × X × R by h(x, y, t) = t(f(x) − f(y)). ˜ M Also deﬁne a map on the tangent bundle TM of M by h : TM → R , (x, v) 7→ Dxf(v). Since dim X × X × R = 2n + 1, dim TM = 2n and M > 2n + 1, Sard’s theorem (2.2.1) implies there is some v ∈ RM lying outside h(X × X × R) and h˜(TM). Set H = Span{v}⊥ M and let projv : R → H be the orthogonal projection. We claim g = projv ◦f : M →

38 3.2 Embedding Theorems 3 Transversality and Embeddings

H is an injective immersion. To see this, take x, y ∈ M such that g(x) = g(y). Then

projv(f(x)) = projv(f(y)) so f(x) − f(y) ∈ ker projv, or in other words, f(x) − f(y) = tv for some t ∈ R. If x 6= y, then t 6= 0 but then we have h(x, y, 1/t) = v, contradicting v 6∈ im h. Thus g is injective. On the other hand, suppose w ∈ ker Dpg. Then by the chain rule, (projv ◦Dpf)(w) = (Dp projv ◦Dpf)(w) = Dp(projv ◦f)(w) = Dpg(w) = 0, so Dpf(w) ∈ ker projv. This implies Dpf(w) = tv for some t ∈ R as above, so we see that h˜(p, t−1w) = v, contradicting v 6∈ im h˜. Hence g is an immersion.

Corollary 3.2.2. Every compact manifold M of dimension n admits an embedding M,→ R2n+1. Proof. This follows from the fact that an injective immersion of a compact manifold is an embedding. To extend this to all smooth manifolds, we introduce the notion of a partition of unity, which will be useful in the future.

Deﬁnition. Let M be a smooth manifold and {Ui}i∈I an atlas of coordinate charts on M. A partition of unity on M is a collection of smooth functions {fi : M → R}i∈I such that

(a) Each fi is identically 0 outside a chart Ui. In other words, fi is supported on Ui.

(b) fi ≥ 0 for all i.

(d) The atlas {Ui}i∈I is locally ﬁnite: each point x ∈ M lies in a neighborhood intersecting a ﬁnite number of the Ui. P (e) For all x ∈ M, i∈I fi(x) = 1. Theorem 3.2.3. Partitions of unity exist for every smooth manifold M.

Proof. We prove this for M = Rn. The general case is obtained by composing with coordinate n charts. For a locally finite atlas of charts {Ui} whose union is R , there exist compact sets ∞ S∞ n {Kj}j=1 with j=1 Kj = R and Kj ⊆ Int(Kj+1). Cover K1 with a finite number of open n n metric balls B1,...,Bk in R and define functions g1, . . . , gk : R → R as follows. First, define the function q : R → R by

( 2 2 e−1/(x−1) e−1/(x+1) , |x| < 1 q(x) = 0, |x| ≥ 1.

n Then q is smooth (check it!) and if B(0, ε) ⊆ B`, then the function h` : R → R deﬁned by x1 xn h`(x1, . . . , xn) = q ε ··· q ε is supported on B`. Now for each 1 ≤ ` ≤ k, set h (x) g (x) = ` . ` Pk r=1 hr(x)

By construction, each g` is smooth, g` = 1 on B` and 0 elsewhere.

39 3.2 Embedding Theorems 3 Transversality and Embeddings

To extend this to K2, notice that K2 r Int(K1) is a compact set, so it is covered by

ﬁnitely many open balls Bk+1,...,Bk+k2 . Repeating the construction above now for the

collection B1,...,Bk,Bk+1,...,Bk+k2 gives us functions g1, . . . , gk, gk+1, . . . , gk+k2 such that for 1 ≤ ` ≤ k, g` is the same function as above and for k + 1 ≤ ` ≤ k + k2, g` is smooth, g` = 1 on B` and 0 elsewhere. Continuing this procedure inductively deﬁnes a partition of P∞ unity fi by restricting `=1 g` to Ui. Theorem 3.2.4 (Whitney’s Embedding Theorem). Every smooth manifold M of dimension n admits an embedding M,→ R2n+1.

Proof. Fix an atlas of charts {Ui} for M and a partition of unity {fi} subordinate to {Ui} and deﬁne a map

h : M −→ R ∞ X x 7−→ ifi(x). i=1 Note that every preimage under h of a compact set (a closed interval) in R is compact – such 1 −1 ε ε a map is called proper. For some fixed small ε > 0, say ε = 4 , set Vi = h i − 2 , i + 1 + 2 −1 and Ci = h (i − ε, i + 1 + ε), so that for each i ≥ 1, Vi ⊆ M is open, Ci ⊆ M is compact and Vi ⊆ Ci. Also notice that the C2j−1 are all disjoint from each other, as are the C2j. For 2n+1 each Ci, Corollary 3.2.2 produces a smooth funciton gi : M → R such that gi restricts to an embedding on Ci and is 0 on M r Ci. After composing with the diffeomorphism, 2n+1 2n+1 x R → R , x 7→ 1+|x|2 , we may assume |gi(p)| < 1 for all p ∈ M. Define

∞ ∞ X X 2n+1 2n+1 fodd(x) = g2j−1, feven(x) = g2j and f : M −→ R × R × R j=1 j=1

x 7−→ (fodd(x), feven(x), h(x)).

Since fodd and feven are bounded, the image of f lies inside K ×R for some compact set K ⊂ R2n+1 ×R2n+1. We claim f is an embedding of M onto a closed subspace of R2n+1 ×R2n+1 ×R. If f(x) = f(y), then h(x) = h(y), so that x, y ∈ Vi for some i ≥ 1. If i is odd, fodd is an embedding of Vi so fodd(x) = fodd(y) implies x = y. Likewise if i is even. Moreover, since h is a proper map and fodd, feven are embeddings, it follows that f(M) is closed. Henceforth we may assume M = f(M) ⊆ R2n+1 × R2n+1 × R. Now by repeated use of the argument in the proof of Theorem 3.2.1, there is some ∼ 2n+1 2n+1 2n+1 2n+1 2n+1 hyperplane H = R in R × R × R such that projH : R × R × R → H is an injective immersion on M. Since Sard’s theorem was used in that proof, we may also 2n+1 2n+1 2n+1 2n+1 assume projH is orthogonal to ker h; that is, if π : R × R × R → R × R is the standard coordinate projection, then ker projH × ker π = {0}. Finally, this implies that 2n+1 projH is in fact an embedding of M as a closed manifold in R . Using more advanced methods, Whitney was able to prove: Theorem 3.2.5 (Whitney Embedding Theorem, Part II). Every smooth manifold M of dimension n admits an immersion M,→ R2n−1 and an embedding M,→ R2n.

40 4 Vector Bundles

4 Vector Bundles

4.1 Vector and Tangent Bundles

Definition. A vector bundle consists of the following data: a smooth manifold B, called the base, a smooth manifold E, called the total space, and a smooth map π : E → B such that for all b ∈ B, the fibre π−1(b) has a vector space structure. Moreover, π must satisfy the “local triviality condition”: for any point b ∈ B, there exists a neighborhood U ⊂ B of b and a diffeomorphism ϕ making the following diagram commute for some Rn: ϕ −1 U × Rn π (U)

proj π

n −1 and for all x ∈ U, ϕ|{x}×Rn : {x}×R → π (x) is a vector space isomorphism. The number n is called the dimension of the vector bundle at b. Remark. If B is connected then the dimension of any vector bundle over B is defined globally. Vector bundles are used in modern differential topology to distinguish between smooth manifolds. We next define the notion of equivalence of vector bundles.

Deﬁnition. Given two vector bundles π1 : E1 → B and π2 : E2 → B over the same base B, we say (E1, B, π1) and (E2, B, π2) are isomorphic vector bundles if there exists a −1 −1 diﬀeomorphism ϕ : E1 → E2 such that for all b ∈ B, ϕ|π−1 (b): π1 (b) → π2 (b) is a vector ∼ 1 space isomorphism. More succinctly, (E1, B, π1) = (E2, B, π2) if and only if the following diagram commutes: ϕ E1 E2

π1 π2

Examples.

1 The trivial bundle of dimension n over any manifold B consists of E = B × Rn and the n −1 n ∼ n projection map π : B × R → B, (b, v) 7→ b. For each b ∈ B, π (b) = {b} × R = R . The local triviality condition in the deﬁnition of a vector bundle intuitively says that every vector bundle locally looks like the trivial bundle.

2 Perhaps the most important vector bundle is the tangent bundle. Let M n ⊂ Rk be a smooth manifold and deﬁne the subset TM ⊂ Rk × Rk by k k TM = {(x, v) ∈ R × R | x ∈ M, v ∈ TxM}.

41 4.1 Vector and Tangent Bundles 4 Vector Bundles

The projection π : TM → M, (p, v) 7→ p, gives a vector bundle structure on TM. −1 Notice that for each x ∈ M, π (x) = TxM so the tangent bundle carries all of the information of the tangent space structures on all of M. Proposition 4.1.1. For a smooth manifold M of dimension n, the tangent bundle TM is a smooth manifold of dimension 2n. Deﬁnition. A one-dimensional vector bundle is called a line bundle. Examples. 3 We claim that the tangent bundle TS1 to the unit circle is isomorphic to the trivial bundle over S1. To show this, we must construct a diﬀeomorphism ϕ : S1 × R → TS1 making the diagram commute:

ϕ S1 × R TS1

proj π

1 1 1 Given any point p = (x, y) ∈ S , TpS is spanned by (−y, x) ∈ TpS . Deﬁne the map

1 1 2 2 ϕ : S × R −→ TS ⊂ R × R ((x, y), t) 7−→ ((x, y), t(−y, x)).

This is clearly smooth and one-to-one, and restricts on each {p}×R to a diﬀeomorphism 1 1 with TpS . Therefore TS is the trivial bundle.

4 One may be tempted to think that all vector bundles are isomorphic to the trivial bundle. In this example we will construct a nontrivial, one-dimensional bundle over projective n-space which is called the tautological bundle. View Sn ⊂ Rn+1 and RP n = Sn/ ∼ where ∼ is the antipodal relation. If x ∈ Sn, denote its corresponding point in projective space by x∗. Deﬁne a subset E ⊂ RP n × Rn+1 by ∗ n n+1 n E = {(x , tx) ∈ RP × R | x ∈ S , t ∈ R}.

Define π : E → RP n by π(x∗, tx) 7→ x∗. Then for each point x∗ ∈ RP n, x∗ may be viewed as a line in Rn+1 through the origin, so π−1(x∗) consists of all vectors in that line defined by x∗. 1 ∼ 1 In the case n = 1, RP = S and E looks like a quotient of an infinite rectangle:

42 4.1 Vector and Tangent Bundles 4 Vector Bundles

Then E is homeomorphic to the Möbiusband. The line bundle E → S1 cannot be the trivial bundle over S1 because E is not diffeomorphic to S1 × R (e.g. S1 × R is orientable and the Möbiusband is not). Hence we have exhibited a nontrivial vector 1 ∼ 1 n bundle over RP = S . In general, the tautological bundle π : E → RP is nontrivial. Definition. A manifold M whose tangent bundle TM is isomorphic to the trivial bundle is called parallelizable.

In Example 3 we showed that S1 is parallelizable. It turns out that for n ≥ 2, Sn is not parallelizable. We will prove this in Section 4.2.

Remark. Invariants of vector bundles give rise to invariants of manifolds by looking at invariants of the tangent bundle π : TM → M. This is related to a rich theory of characteristic classes, notably developed by Milnor and Stasheﬀ.

Let π : E → B be a vector bundle over B and consider two overlapping charts Ui,Uj ⊂ B over each of which the bundle is trivial. We’d like to describe what happens in their intersection. Consider the diagram

ϕ ϕj n i −1 n (Ui ∩ Uj) × R π (Ui ∩ Uj) (Ui ∩ Uj) × R

proj π proj

U1 ∩ U2

For any x ∈ Ui ∩ Uj, we have linear isomorphisms

n ϕi −1 ϕj n {x} × R −→ π (x) ←−{x} × R .

−1 Then ϕij := ϕj ◦ ϕi is a vector space isomorphism. In other words, at each x ∈ B there n n is a family of vector space isomorphisms ϕij : R → R parametrized by the charts Ui,Uj containing x. Viewing this as a family of functions on patches of B over which we may vary x, we instead get maps

ψij : Ui ∩ Uj −→ GLn(R)

x 7−→ ϕij|{x}×Rn . So there is a correspondence between vector bundles of dimension n over B and parametrized families of maps from B into GLn(R).

Example 4.1.2. Constructing the trivial bundle this way produces maps ψij : Ui ∩ Uj → GLn(R) taking x 7→ I for all x ∈ Ui ∩ Uj. 1 ∼ 1 Example 4.1.3. Consider the tautological bundle E → RP = S over the circle. Let’s ∗ instead view this bundle as a family of maps into GL1(R) = R r {0} = R . Take two overlapping charts Ui and Uj:

43 4.2 Sections 4 Vector Bundles

Ui ) ) 1

) S )

−1 ∼ −1 ∼ Then π (Ui) = Ui ×R and π (Uj) = Uj ×R, and Ui ∩Uj = A∪B, the union of two disjoint ∗ open intervals on the circle. We can deﬁne the map ψij : Ui ∩ Uj → R by our choices of sending A and B each to either R>0 or R<0. Thus the equivalence classes of line bundles 1 over S are determined by whether the maps ψij send the disjoint pieces of Ui ∩ Uj into the same connected component of R∗ or the opposite components. Hence the only line bundles over the circle are, up to isomorphism, the tautological and trivial bundles.

4.2 Sections

Deﬁnition. Given a vector bundle π : E → B, a section of the bundle is a smooth map s : B → E such that π ◦ s = idB. When E = TM is the tangent bundle over a manifold M, a section s : M → TM is instead called a smooth tangent vector ﬁeld on M.

Example 4.2.1. The trivial section s : B → E, b 7→ 0 ∈ π−1(b) is smooth for any vector bundle π : E → B. In particular, sections exist for every vector bundle over B. In fact, it turns out that the space of sections for a particular bundle is uncountable!

Question. Given a bundle π : E → B, is there a section s : B → E that is nonvanishing on B? That is, does there exist a section such that s(b) 6= 0 for all b ∈ B?

Example 4.2.2. There is always a nonvanishing section on any trivial bundle B × Rn → B, for n ≥ 1, since for a fixed nonzero vector v ∈ Rn, s(b) = (b, v) is smooth. 1 ∼ 1 Example 4.2.3. Let E be the tautological bundle over RP = S and recall that E may be viewed as an infinite Möbiusband. This is depicted below with 0 marked as the ‘equator’.

Since a section must be smooth, by the intermediate value theorem its graph in E must cross the ‘equator’ at some point, and hence it must vanish somewhere. (This is illustrated in red in the ﬁgure.) Therefore there are no nonvanishing sections on the tautological bundle. On the other hand, the tangent bundle TS1 over the circle has various nonvanishing sections, which are all tangent vector ﬁelds such as the one pictured below.

44 4.2 Sections 4 Vector Bundles

F (x, y) = (−y, x)

The so-called ‘hairy ball theorem’, stated below, states the same phenomenon does not happen with the 2-sphere.

Theorem 4.2.4 (Hairy Ball Theorem). There is no nonvanishing smooth vector ﬁeld on S2.

Corollary 4.2.5. S2 is not parallelizable.

Proof. By the hairy ball theorem, the tangent bundle TS2 is not isomorphic to the trivial bundle over S2, since the latter always has a nonvanishing section. We next prove that none of the other even spheres S2k are parallelizable. To do so, we ﬁrst need

Lemma 4.2.6. If Sn admits a nowhere vanishing vector ﬁeld then the antipodal map a : Sn → Sn is nullhomotopic.

Proof. Given a smooth nonvanishing vector ﬁeld v : Sn → Rn+1, we know that for all x ∈ Sn, v(x) · x = 0. Since v(x) 6= 0 for any x ∈ Sn, we may assume |v(x)| = 1 everywhere. This allows us to deﬁne a smooth map F : Sn × [0, 1] → Sn by F (x, t) = x cos(πt) + v(x) sin(πt). Notice that

|F (x, t)|2 = |x|2 cos2(πt) + 2x · v(x) cos(πt) sin(πt) + |v(x)|2 sin2(πt) = cos2(πt) + 0 + sin2(πt) = 1.

Thus the image of F lies in Sn. Also, F (x, 0) = x and F (x, 1) = −x, so F is a smooth homotopy between the identity and the antipodal map on Sn.

Theorem 4.2.7. For n ≥ 2 even, Sn is not parallelizable.

Proof. We saw in Section 2.5 that the antipodal map a has degree (−1)n+1, so when n is even, deg a = −1. Since Brouwer degree is preserved by homotopy, this shows that a is not homotopic to the identity. Hence by Lemma 4.2.6, Sn does not admit a nonvanishing vector ﬁeld when n is even.

45 4.2 Sections 4 Vector Bundles

Example 4.2.8. Consider S3 ⊂ R4. As with S1 and S2, the outward-pointing unit normal 3 to any point (x1, x2, x3, x4) ∈ S is just hx1, x2, x3, x4i. Consider the following tangent vector ﬁelds on S3, constructed from vectors orthogonal to the normal vector:

~v1 = h−x2, x1, −x4, x3i, ~v2 = h−x3, x4, x1, −x2i, ~v3 = h−x4, −x3, x2, x1i.

3 These are all nonvanishing, so TS admits three nonvanishing sections ~v1,~v2,~v3 at each point x ∈ S3. Moreover, these three sections are linearly independent.

Deﬁnition. Two sections s1, s2 of a vector bundle π : E → B are said to be linearly independent if s1(x) and s2(x) are linearly independent for every x ∈ B. An even more interesting question one can ask is: given an n-dimensional vector bundle, what is the maximum number of linearly independent sections it can have? We saw in the last example that the tangent bundle to S3 admits three linearly independent sections. Intu- itively, the maximum number of linearly independent sections is a measure of how nontrivial the bundle is. Characteristic classes are a type of algebraic invariant which correspond to obstructions to the existence of k linearly independent sections of a vector bundle. Theorem 4.2.9. An n-dimensional vector bundle is trivial if and only if it admits n linearly independent sections.

n n Proof. ( =⇒ ) The trivial bundle B × R → B admits sections si : B → B × R , x 7→ (x, ei), n where ei is the ith standard basis vector in R . These are clearly linearly independent at every point in the base. ( ⇒ = ) Conversely, suppose E −→π B is an n-dimensional bundle with linearly independent sections s1, . . . , sn. Deﬁne a map

n ψ : B × R −→ E n X (x, v1, . . . , vn) 7−→ visi(x). i=1

Clearly ψ is a vector space isomorphism on each ﬁbre of the bundle, since the si(x) are linearly independent, and the following diagram commutes: ψ B × Rn E

Thus E is trivial. Corollary 4.2.10. S3 is parallelizable. This result about the 3-sphere is tied to the very special fact in 3-manifold theory that any closed, oriented 3-manifold is parallelizable. One may be tempted to conjecture that all odd-dimensional spheres are parallelizable, but it turns out that the only parallelizable spheres are S1,S3 and S7.

46 4.3 Constructing New Vector Bundles 4 Vector Bundles

Deﬁnition. A Lie group is a smooth manifold G with a group structure such that the multiplication map G × G → G and the inverse map G → G are smooth. Example 4.2.11. The most important class of Lie groups arise as general linear groups GLn(R) and their subgroups. These are called the classical Lie groups.

Definition. For a Lie group G, the tangent space to the identity, TeG, is called the Lie algebra of G. It is denoted g. There is a rich theory of abstract Lie algebras which arise from putting a certain nonas- sociative structure on an algebra over a field. The classical theory however revolves around Lie groups and their Lie algebras. Proposition 4.2.12. Every Lie group is parallelizable. Proof. Let G be a Lie group and let n be the dimension of G as a manifold. Then the tangent bundle over G is a vector bundle of dimension n. We will prove that TG → G admits n linearly-independent sections everywhere. Consider g = TeG, the Lie algebra of G. ∼ n Since g is an n-dimensional vector space, there is a choice of basis of g giving g = R , say {v1, . . . , vn}. We will show that for each i, the map si(g) = (g, g · vi) ∈ TG is a vector field and that g · v1, . . . , g · vn are linearly independent, for some appropriate action of g on the tangent vectors at the identity. Since G is a Lie group, left multiplication by g is a smooth map λg : G → G for each g ∈ G. Note that λg−1 is also smooth and clearly λg ◦ λg−1 = λg−1 λg = idG. So λg is a diffeomorphism. Moreover, this induces a differential at any x ∈ G, namely dxλg : TxG → TgxG, which is a vector space isomorphism since it comes from a diffeomorphism. In particular, for x = e this gives an isomorphism deλg : g → TgG. Set Λg = deλg. Then we n define the action of G on g by g · v = Λg(v). Define {si : G → TG}i=1 as suggested above: si(g) = (g, g · vi) = (g, Λg(vi)). Since each Λg is a vector space isomorphism and {v1, . . . , vn} was a basis for g, {Λg(v1),..., Λg(vn)} is a linearly independent set of vectors. Finally, each si is a vector field since the action of G on g is smooth. This proves G is parallelizable.

4.3 Constructing New Vector Bundles

π1 π2 Deﬁnition. Given two vector bundles E1 −→ B1 and E2 −→ B2, their product bundle is the vector bundle π1 × π2 : E1 × E2 −→ B1 × B2. Lemma 4.3.1. Given two manifolds M and N, the tangent bundle T (M ×N) is isomorphic to TM × TN as vector bundles. Example 4.3.2. Consider the torus X ∼= S1 × S1. The tangent bundle to the torus is a product bundle TX = TS1 × TS1. Since S1 is parallelizable, TS1 is trivial and therefore so is TS1 × TS1, so the torus is also parallelizable.

π2 Deﬁnition. If E −→ B2 is a vector bundle and f : B1 → B2 is a smooth map, then we π2 deﬁne the induced bundle, also called the pullback of E −→ B2 along f, to be the vector ∗ π1 bundle f E −→ B1 making the following diagram commute:

47 4.3 Constructing New Vector Bundles 4 Vector Bundles

f ∗E E

π1 π2 f B1 B2

∗ Formally, f E = {(b, e) ∈ B1 × E | f(b) = π2(e)} ⊆ B1 × E and the dashed arrows are the ∗ canonical projections B1 × E → B1 and B1 × E → E restricted to f E.

π2 Proposition 4.3.3. For every vector bundle E −→ B2 and smooth map f : B1 → B2, the ∗ π1 pullback f E −→ B1 is a vector bundle.

−1 −1 Proof. For each point x ∈ B1, π1 (x) = {(x, e) | f(x) = π2(e)} = {x} × π2 (x), so each ﬁbre in the pullback is a vector space. The remaining axioms are easily veriﬁed. Pullbacks are extremely important for the following reason:

Theorem 4.3.4. There exists a vector bundle Euniv → Buniv, called the universal bundle, such that every bundle E → B is a pullback of Euniv along some smooth map f : B → Buniv.

π1 π2 Deﬁnition. Given two vector bundles E1 −→ B and E2 −→ B over the same base space, the Whitney sum of the bundles is the vector bundle

π1⊕π2 E1 ⊕ E2 −−−→ B

deﬁned as the pullback along the diagonal map of the product bundle:

∗ E1 ⊕ E2 = ∆ (E1 × E2) E1 × E2

π1 ⊕ π2 π1 × π2 ∆ B B × B x (x, x)

Deﬁnition. Suppose M m ⊆ Rn is a smooth manifold and S ⊂ M is a submanifold of dimension k. The normal bundle over S in M consists of the total space

n VS,M := {(x, v) ∈ S × R : v ∈ TxM and v · w = 0 for all w ∈ TxS}

n and the projection π : VS,M ⊆ S × R → S. Proposition 4.3.5. Given a submanifold S ⊂ M, the tangent and normal bundles over S ∼ satisfy TS ⊕ VS,M = TM|S, where ⊕ is the Whitney sum and TM|S is the restriction of the tangent bundle over M to S.

Theorem 4.3.6. Given any manifold M ⊆ Rk and a submanifold S ⊂ M, for suﬃciently small ε > 0, the ε-neighborhood of S in M (a tubular neighborhood) is diﬀeomorphic to the total space VS,M of the normal bundle of S in M.

48 4.3 Constructing New Vector Bundles 4 Vector Bundles

Categorically, there are many ways of obtaining new vector spaces from old ones. These operations are called functors. Examples of functors on vector spaces include: direct sum (⊕), tensor product (⊗), exterior power (Vk) and the vector space dual (V ∗). It turns out that these examples are smooth functors. It is a general principle that any smooth functor of vector spaces determines an analagous construction of vector bundles. For example, recall the vector space dual of V is deﬁned by

∗ V = Hom(V, R) = {f : V → R | f is linear}.

∗ Given a vector bundle E −→π B, we can construct the dual bundle E∗ −→π B. In light of our description at the end of Section 4.1, regard E −→π B as a family of smooth maps ϕij : Ui ∩ Uj → GLn(R), where {Ui}i∈I cover B. Taking transposes gives us smooth maps ∗ ϕij : Ui ∩ Uj −→ GLn(R) ∗ T x 7−→ ϕij(x) := ϕij(x) . Per the equivalence of such families of maps and vector bundles, we obtain a new vector bundle: ∗ Definition. Given a vector bundle E −→π B, its dual bundle is the vector bundle E∗ −→π B such that each fibre is the vector space dual: (π∗)−1(x) = (π−1(x))∗. The structure of the ∗ ∗ total space E is determined by the transition maps ϕij defined above. Notice that it’s not enough to just define the fibres (π∗)−1(x) = (π−1(x))∗, as there is no canonical isomorphism between a vector space V and its dual V ∗. For this reason one requires the extra information carried by the maps ϕij : Ui ∩ Uj → GLn(R). Similarly, suppose we have two vector bundles over the same base space:

E1 E2

π1 π2 B

Then we can construct the tensor bundle E1 ⊗ E2 → B using the same procedure as for π1 the dual bundle. View E1 −→ B as a family of smooth maps ϕij : Ui ∩ Uj → GLm(R) and π2 likewise view E2 −→ B as ψk` : Vk ∩ V` → GLn(R). By intersecting the charts Ui,Vk if necessary, we may assume {Ui} = {Vk} as sets. Then there is a smooth map for each pair of charts Ui,Uj ⊂ B given by tensoring matrices:

Φij : Ui ∩ Uj −→ GLmn(R) x 7−→ ϕij(x) ⊗ ψij(x).

If E1 → B is a bundle of dimension m and E2 → B is dimension n, as suggested by the notation, then E1 ⊗ E2 → B will be a bundle of dimension mn.

π1 π2 Deﬁnition. Given two vector bundles E1 −→ B and E2 −→ B, the tensor bundle of these π1⊗π2 −1 −1 −1 bundles is E1 ⊗ E2 −−−→ B, where for each x ∈ B, (π1 ⊗ π2) (x) = π1 (x) ⊗ π2 (x) as vector spaces. The structure of the total space E1 ⊗ E2 is determined by the transition maps Φij deﬁned above.

49 4.3 Constructing New Vector Bundles 4 Vector Bundles

Example 4.3.7. Consider the tautological line bundle E −→π S1 over the circle. There is a tensor bundle E ⊗ E → S1 which by the description above must also be a line bundle. It turns out that E ⊗ E → S1 is (isomorphic to) the trivial line bundle over S1. This shows that there is a group structure on the isomorphism classes of line bundles over S1. In general, one can put a group structure on the set of isomorphism classes of line bundles over any base B, with the group law being E1 · E2 := E1 ⊗ E2 and inverses given by E−1 := E∗, the dual bundle. In the example of line bundles over S1, the group of line bundles is isomorphic to Z/2Z, with the tautological bundle the nontrivial element. For a vector bundle E → B, one can construct the kth exterior power of the bundle, VkE → B, in a similar fashion as the tensor bundle. An important application of this construction is to the theory of diﬀerential forms.

Deﬁnition. For a manifold M, the dual bundle of the tangent bundle TM over M is called the cotangent bundle over M, denoted T ∗M.

Deﬁnition. A section of the kth exterior power of the cotangent bundle, VkT ∗M, is called a diﬀerential k-form on M.

Example 4.3.8. In calculus, consider the antiderivative of a single-variable function, R f(x) dx. Here, dx is an example of a 1-form on R.