Cobordism and Hopf’s Theorem
E.J. Sanchez
1
Introduction
In this paper we will provide an exposition of some of Milnor’s Topology 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 manifold
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 topological 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 manifolds 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 manifolds 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) = {v 1 (y), . . . v m−n (y)} where the v i (y) form a basis of
Ty N ⊥ ⊂ Ty M
To see a straightforward example of a framing, take the n-sphere Sn embedded 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 framing comes up somewhat naturally. If
we choose a regular value y ∈ N in
the image of f , then we have that
Tx f −1 (y) is defined by the vanishing of the differential dx f , for any x ∈ M in
2
the fiber f −1 (y). Thus, since dx f 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
value y ∈ Sp . Let υ = (v1 , . . . vp ) be a positively oriented basis for the tangent
space Ty Sp , and observe by what was stated above that we have the natural
isomorphism Ty Sp ∼
= Tx⊥ f −1 (y). Thus υ induces a framing of the submanifold
−1
f (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.
Lemma 2.1. Let υ and υ 0 be positively oriented bases of Tx⊥ f −1 (y). Then
(f −1 (y), υ) and (f −1 (y), υ 0 ) 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
bases of Tx⊥ f −1 (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 υ 0 . 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 sufficiently 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
family of rotations rt : Sp −→ Sp so that r1 (y) = z and the following are
satisfied:
• r0 = id,
• rt = r1 for 0 < t ≤ 1,
• ∀t, rt−1 (z) lies on the great circle from y to z.
We may now define a homotopy F : M × I −→ Sp by F (x, t) = rt · f (x).
For each t, z is a regular value of the composition rt ◦ f : M −→ Sp , 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
cobordism between f −1 (z) and (r1 ◦ f )−1 (z) = f −1 ◦ r1−1 (z) = f −1 (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
to r1 ◦ f and hence by Lemma 2.3 we have that f −1 (y) is framed cobordant
to (r1 ◦ f )−1 (y) = f −1 ◦ r1−1 (y) = f −1 (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 implication 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 Euclidean space, the construction of this neighborhood uses the framing to set up the map g(x, t1 , . . . , tp ) = x + t1 v1 (x) +
. . . tp vp (x) from N × Rp into M . One may observe that on some small neighborhood 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
N × B (0). Scaling this -neighborhood to all of Rp is straightforward, and
the proof holds. If M is not Euclidean space then replace the straight lines
ti vi (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:
H(x, t) =
f (x)
h−1 [t · h(f (x) + (1 − t) · h(g(x))]
if x ∈ V
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 point ȳ of y, while
not disturbing the diffeomorphism relation.
We thus have maps f, g : V −→ Sp \ ȳ, which (in view of the diffeomorphisms) we may describe as maps F, G : N × Rp −→ Rp which satisfy:
• F −1 (0) = G−1 (0) = N × 0
• dF(x,0) = dG(x,0) = projection to Rp
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
smooth map λ : Rp −→ 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 (seemingly 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
basis of Tx M at each point x in the fiber. So, if for x ∈ f −1 (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
−1
see
P that the framed cobordism class of f (y) is uniquely determined by
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.
8