Cobordism and Hopf’s Theorem

E.J. Sanchez

1 Introduction

In this paper we will provide an exposition of some of Milnor’s from the Differentiable Viewpoint, introducing the theory of cobordism and using it to prove the following theorem due to Heinz Hopf:

Theorem 1 (Hopf’s Theorem). Let M be a connected, oriented without boundary of dimension p, and let f,g : M −→ Sp be smooth maps. Then f and g are homotopic ⇔ deg(f) = deg(g).

Before proving this theorem, we will lay some groundwork with a brief overview of cobordism, framed cobordism, and the Pontryagin manifold. We will then prove Theorem 2, which connects the Pontryagin manifold (a geometric construction) with the homotopy class of the map (a topolog- ical notion) and will be the most important ingredient in proving Hopf’s Theorem:

Theorem 2. With assumptions as in Hopf’s Theorem, f,g : M −→ Sp are homotopic ⇔ the associated Pontryagin are framed cobordant.

2 Cobordism and Framed Cobordism

In this section we define the notions of cobordism and framed cobordism, and give a few basic examples.

Definition 1. Let N 0 and N be manifolds without boundary, of dimension p, embedded in some ambient manifold M. Then we say a submanifold X of M × I is a cobordism of N 0 and N if ∂X = N 0 × {0} ∪ N × {1}.

1 Thus, cobordism forms an equivalence relation on the set of smooth mani- folds without boundary. To see this, note that reflexivity is clear if we simply choose X = N × I. Symmetry also follows immediately from the definition. Finally, transitivity holds since if X1 is a cobordism of M and N, and X2 is a cobordism of N and P , then we can rescale each by one half and then glue X1 and X2 along the boundary manifold N to get a manifold with boundary equal to M × {0} ∪ P × {1}. To utilize the theory of cobordism, we place additional structure on our manifolds, namely the framing of a submanifold. Note that in the following definition and in the rest of the paper, we will denote the dimension of manifolds M and N by m and n, respectively. Definition 2. A framing of the submanifold N ⊂ M is a smooth function v which takes each y ∈ N and assigns to it a basis of the normal vectors to N in M at y. That is, v(y) = {v1(y), . . . vm−n(y)} where the vi(y) form a basis of ⊥ TyN ⊂ TyM To see a straightforward example of a framing, take the n-sphere Sn em- bedded in Rn+1. Then there is the natural framing which assigns to each point x ∈ Sn the unit normal vector to x, i.e. thought of as one unit outward along the line from y to the origin. The normal tangent space has dimension equal to one, so each unit vector forms a basis. Given framed submanifolds N and N 0 in M, we can naturally expand the definition of cobordism by introducing the notion of framed cobordism: N and N 0 are framed cobordant if there is a cobordism X ⊂ M, where X itself is a framed submanifold of M that on N and N 0 restrics to the given framings on those submanifolds. Now, given a map f : M −→ N, observe that this notion of a fram- ing comes up somewhat naturally. If we choose a regular value y ∈ N in the image of f, then we have that −1 Txf (y) is defined by the vanishing of the differential dxf, for any x ∈ M in

2 −1 the fiber f (y). Thus, since dxf is a linear map between vector spaces, the subspace spanned by the tangent vectors normal to f −1(y) at x is isomorphic to the tangent space of Imf at y. We then consider the particular case of f : M −→ Sp, and select a regular p value y ∈ S . Let υ = (v1, . . . vp) be a positively oriented basis for the tangent p space TyS , and observe by what was stated above that we have the natural p ∼ ⊥ −1 isomorphism TyS = Tx f (y). Thus υ induces a framing of the submanifold f −1(y). This submanifold f −1(y) ⊂ M with the induced framing (denoted f ∗υ) is called the Pontryagin manifold associated to f at y. The following theorem justifies the use of the article “the” Pontryagin manifold associated to f; in fact the Pontryagin manifold is unique up to framed cobordism:

Theorem 3. The Pontryagin manifolds associated to f at any two regular values are framed cobordant.

Proving this statement requires a few preparatory results however, the first of which shows that any two positively oriented framings of f −1(y) yield framed cobordant submanifolds.

0 ⊥ −1 Lemma 2.1. Let υ and υ be positively oriented bases of Tx f (y). Then (f −1(y), υ) and (f −1(y), υ0) are framed cobordant.

Proof. Note that we have the obvious cobordism f −1(y) × I, so we simply need to find compatible framing. But the space of all positively oriented ⊥ −1 bases of Tx f (y) can be identified with the subspace of matrices in GLn(R) that have positive determinant, and since this is a connected space there is a path connecting υ and υ0. Each point of this path represents a positively oriented basis which varies continuously, so this gives the compatible framing as desired.

The next result shows that if regular values are sufficiently close together, then their inverse images are frame cobordant.

Lemma 2.2. Given a regular value y of f in Sp, if we choose any z suffi- ciently close to y, then f −1(y) is framed cobordant to f −1(z).

Proof. First observe that the set of critical values, denoted f(C), in Sp is compact, so we may choose some small  > 0 so that the closed -ball around

3 y contains only regular values. Now, given z in this -ball, choose a smooth p p family of rotations rt : S −→ S so that r1(y) = z and the following are satisfied:

• r0 = id,

• rt = r1 for 0 < t ≤ 1,

−1 • ∀t, rt (z) lies on the great circle from y to z.

p We may now define a homotopy F : M × I −→ S by F (x, t) = rt · f(x). p For each t, z is a regular value of the composition rt ◦ f : M −→ S , and so z is also a regular value of the homotopy F . Therefore, F −1(z) is a framed submanifold of M × I and gives a framed −1 −1 −1 −1 −1 cobordism between f (z) and (r1 ◦ f) (z) = f ◦ r1 (z) = f (y), thus proving the lemma.

Lemma 2.3. If f and g are smooth homotopic and y is a regular value for both, then f −1(y) is framed cobordant to g−1(y).

Proof. Let F be a homotopy from f to g, and choose a regular value z for F which is sufficiently close to y so that both f −1(y) and g−1(y) are framed cobordant to f −1(z) and g−1(z) respectively. F −1(z) then serves as a framed cobordism for f −1(z) and g−1(z), and the claim follows from Lemma 2.2.

Now, given a map f : M −→ Sp, we would like to finish showing that the Pontryagin manifold f −1(y) of f is independent (up to framed cobordism) of the regular value y chosen. So we suppose two regular values y and z of f, and note that we may choose rotations rt, just as in Lemma 2.2 of the sphere so that r0 = id and r1(y) = z. Then F (x, t) = rt ◦ f is a homotopy from f −1 to r1 ◦ f and hence by Lemma 2.3 we have that f (y) is framed cobordant −1 −1 −1 −1 to (r1 ◦ f) (y) = f ◦ r1 (y) = f (z), thus proving Theorem 3. So we are justified in referring to “the” Pontryagin manifold of a map f : M −→ Sp, up to framed cobordism class. But in fact much more is true: Any compact framed submanifold N of codimension p in a manifold M is the Pontryagin manifold of some mapping f : M −→ Sp. We will not prove this here, but the idea is to find an appropriate “product” neighborhood of N (in a sense which will be defined later) and compose maps from this neighborhood to Rp with an embedding

4 Rp −→ Sp. This is an interesting fact which nicely rounds out the basic theory of Pontryagin manifolds, and will be needed to prove the final statement. Returning to our immediate goal of proving Theorem 2, we observe that Lemma 2.3 finishes showing the forward direction; we now proceed to argue for the reverse direction, which will suffice to show the more difficult impli- cation in Hopf’s Theorem. To begin this, we will first argue for a slightly weaker statement, where the Pontryagin manifolds are assumed to be strictly equal rather than merely framed cobordant. But in fact most of the work consists of proving this weaker statement.

Lemma 2.4. If the framed manifold (f −1(y), f ∗υ) is equal to (g−1(y), g∗υ), then f is smoothly homotopic to g.

The proof of this statement will require an auxiliary result, for which we merely sketch a brief argument.

Lemma 2.5 (Product Neighborhood Theorem). Let N be a framed embedded submanifold of M of codimension p. Then there exists a neighborhood of N in M that is diffeomorphic to the product N × Rp. Moreover, we may choose the diffeomorphism such that each x ∈ N is mapped to (x, 0) in N × Rp, and so that each normal frame υ(x) corresponds to the standard basis for {x} × Rp. In the case where M is , the construction of this neigh- borhood uses the framing to set up the map g(x, t1, . . . , tp) = x + t1v1(x) + p . . . tpvp(x) from N × R into M. One may observe that on some small neigh- borhood of (x, 0) ∈ M × Rp this map g is injective ∀x ∈ N, and from this argue that there exists an  > 0 such that g is a diffeomorphism from p N × B(0). Scaling this -neighborhood to all of R is straightforward, and the proof holds. If M is not Euclidean space then replace the straight lines tivi(x) with geodesics and an analogous argument goes through. Now, with this result in hand, we give the proof of Lemma 2.4. Proof. We assume f −1(y) = g−1(y), and want to show that f is homotopic to g. To simplify notation, we follow Milnor and denote N = f −1(y). We first suppose that there exists some neighborhood V of N such that f = g on V . The claim will follow simply in this case, so most of the argument will be in showing that we may smoothly deform f so that such a neighborhood occurs.

5 So, assuming such a neighborhood V , we let h : Sp \{y} −→ Rp be stereographic projection onto Rp. We may then define a continuous map H : M × I −→ Sp by:

 f(x) if x ∈ V H(x, t) = h−1[t · h(f(x) + (1 − t) · h(g(x))] if x ∈ M \ N which gives a homotopy from g to f as mappings on M. Therefore, it is enough to continuously deform f (though without adding more points to the pre-image of y!) so that it coincides with g on some such neighborhood V . To this end, we may by Lemma 2.5 choose a neighborhood V ⊃ N that is diffeomorphic to N × Rp. Note that we may scale V such that f(V ) and g(V ) each do not contain the antipodal pointy ¯ of y, while not disturbing the diffeomorphism relation. We thus have maps f, g : V −→ Sp \ y¯, which (in view of the diffeomor- phisms) we may describe as maps F,G : N × Rp −→ Rp which satisfy:

• F −1(0) = G−1(0) = N × 0

p • dF(x,0) = dG(x,0) = projection to R

Our goal then is to smoothly deform F , but in doing so we wish to ensure that no new points are added to N; in terms of the properties above, we want no additional points of M to be mapped to 0. We would like to choose a small punctured neighborhood U of 0 ∈ Rp so that for u ∈ U we have F (x, u) · u > 0 and G(x, u) · u > 0; this will imply that F (N × U) and G(N × U are mapped to the same open half-space in Rp and so the straight line homotopy (1 − t)F (x, u) + tG(x, u) will not intersect 0. To do this, note that by Taylor’s theorem there exists a constant c such that kF (x, u) − uk ≤ c kuk2 for kuk ≤ 1, which implies |(F (x, u) − u) · u| ≤ c kuk3, and so F (x, u) · u ≥ kuk2 − c kuk3 > 0.

6 So if we choose  = min{1, c−1}, then for u ∈ B(0, ) \ 0 we will have F (x, u) · u > 0. Since a similar inequality holds for G, taking the smaller of the two neighborhoods gives us our desired U. Finally, to avoid moving distant points, it is worthwhile to choose a p  smooth map λ : R −→ R which is 1 on the closure of B(0, 2 ) and 0 on the (closed) complement of B(0, ). We then have the following homotopy  from F = F0 to a mapping F1 which coincides with G on B(0, 2 ) and has no new zeros:

Ft(x, u) = [1 − λ(u)t]F (x, u) + λ(u)tG(x, u). This homotopy, when translated back into a homotopy between f and g : V −→ Sp, thus suffices to prove the claim.

We are now ready to finish the argument for the reverse implication in Theorem 2, using Lemma 2.4. Suppose we have smooth maps f, g : M −→ Sp with a framed cobordism (X, υ) between f −1(y) and g−1(y). Then there is a smooth map F : M × I −→ Sp such that X is the Pontryagin manifold associated to F ; i.e. X = F −1(y) for some y ∈ Sp and the induced framing is exactly υ. But note that the Pontryagin manifolds of F0 and f are exactly equal, and similarly with F1 and g; by the above lemma this implies that f is homotopic to F0 and g homotopic to F1. Therefore by transitivity f is homotopic to g, and Theorem 2 holds.

3 Hopf’s Theorem

Finally, as an example of the power of this theory of framed cobordism and the Pontryagin manifold, we will see that it makes it possible to give a (seem- ingly simple) proof of the Hopf Theorem, where M is a manifold of dimension p.

Theorem 4 (Hopf Theorem). If M is connected, oriented and boundaryless, then two maps f, g : M −→ Sp are smoothly homotopic if and only if they have the same degree.

Proof. Note that it by Theorem 2, it is enough to show that f and g have the same degree if and only if the Pontryagin manifolds of f and g are framed cobordant. But note that since dim(M) = p, f −1(y) and g−1(y) are merely

7 collections of points, and a framing of each of these consists of a preferred −1 basis of TxM at each point x in the fiber. So, if for x ∈ f (y) the induced basis f ∗υ has positive orientation, then f preserves orientation at that point and say sgn(f, x) = 1; if not then sgn(f, x) = −1. But it is not hard to see that the framed cobordism class of f −1(y) is uniquely determined by P x∈f −1(y)sgn(f, x). Indeed, given a collection of 2n points with half positive and half negative orientations (so that the degree of this map is zero) then we construct a cobordism X of this collection and the empty set by letting X be the disjoint union of n paths through M × I, each of which runs from a point of positive orientation to one of negative orientation. Alternatively, there is a natural framed cobordism between points of the same orientation: the path connecting the two points in M, but viewed as a submanifold in M × I. Putting these two constructions together deals with every case, and so we conclude that the framed cobordism class is uniquely determined by the degree. But a basic fact about the degree of a map says that: P deg(f) = x∈f −1(y)sgn(f, x). Thus we have: deg(f)) = deg(g) ⇔ the Pontryagin manifolds of f and g are framed cobordant ⇔ f and g are homotopic.

And Hopf’s Theorem is proven.

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