hopf degree theorem

7/29/2019 Hopf Degree Theorem

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Chris ZinMTH 864

April 2008

THE HOPF DEGREE THEOREM

Abstract. This paper provides a proof of the Hopf Degree Theorem as statedin Chapter 3, Section 6 of [1]. The proof will be provided through a series ofexercises and hints as laid out in this section, and as such, will we often makeuncited reference to [1]. Also, the hints are riddled with errors so this paperwill assist at clearing up these errors for the reader.

1. Introduction

The Hopf Degree Theorem is quite ambitious. It fulfills the goal of classifyingall maps (up to homotopy) from any compact, connected, oriented k-manifold intoSk by appealing to degree theory. That is to say, the homotopy class of any suchmap is given by a single integer. Remarkable, indeed!

The theorem is quite fundamental to the subject of topology, and a version wasproved by Hopf around 1925 [3, 2.2]. Hence, the theorem derives it name directlyfrom the man who proved it, rather than just being named in honor of him.

Such a powerful theorem undoubtedly comes in handy to all sorts of mathemati-cians. Of course, a theorem is mere conjecture without its proof, so to appease thesemathematicians (and hopefully yourself!) let us plot a course through the turbu-lent ocean of discovery prior to docking at the harbor of the Hopf Degree Theorem.Before embarking, remember to bring your spyglass to catch all the details alongthe way, for as many a globetrotter would say, the voyage is more important thanthe destination. 1

First off, let us establish some useful preliminary results in a section which shouldbe entitled, Better than Sams Preliminaries.

2. Preliminaries

Exercise 1. Let f : U Rk be any smooth map defined on an open subset Uof Rk, and let x be a regular point, with f(x) = z. Let B be a sufficiently smallclosed ball centered at x, and define f : B Rk to be the restriction of f tothe boundary of B. Prove that W(f,z) = +1 if f preserves orientation at x andW(f,z) = 1 if f reverses orientation at x.

Proof. For simplicity, take x = 0 = z, and set A = df0. By Taylors theorem we canwrite f(x) = Ax + (x), where (x)/|x| 0 as |x| 0. Define ft(x) = Ax + t(x)for t [0, 1]. Then, ft is a homotopy of f0(x) = Ax and f1(x) = f(x).

Since x is a regular point, A is surjective, and hence, an isomorphism. So,the image of the unit ball under A contains some closed ball of radius c > 0.Furthermore, the boundary of the unit ball (Sk1) is mapped diffeomorphically

1It should be noted that this part of the introduction is an homage to the writing styleGuillemin and Pollack used throughout their book, as the author of this paper would not normallyuse such odd references.

1


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2 THE HOPF DEGREE THEOREM

onto the boundary of the image, by Exercise 2, Chapter 2, Section 1. Thus, |Ax| > cfor all x Sk1. It then follows, from |A x|x| | > c and by linearity, that |Ax| > c|x|

for all x Rk \ {0}.Since (x)/|x| 0 as |x| 0, we can choose the radius of the ball B small

enough so that |(x)||x| c|x| 1

2c|x| =

1

2c|x|.

Therefore, we can define Ft(x) =ft(x)|ft(x)|

: B I Sk1, a homotopy of F0(x) =Ax|Ax| and F1(x) =

f(x)|f(x)|

. Since they are homotopic, the degrees of these two maps

are equal, which implies W(A, 0) = W(f , 0).Now we invoke the linear isotopy lemma of Chapter 3, Section 4, which is stated

in the appendix. This tells us that A is homotopic to either the identity (if A

preserves orientation) or the reflection map (x1, . . . , xk) (x1, x2, . . . , xk) (if Areverses orientation). So, if A preserves orientation, then W(A, 0) = +1, and if Areverses orientation, then W(A, 0) = 1. The result follows immediately.

Essentially, Exercise 1 limits the way in which local diffeomorphisms can windaround a point. In Exercise 2 we exploit this information to count preimages.

Exercise 2. Letf : B Rk be a smooth map defined on some closed ball B inRk.Suppose that z is a regular value of f that has no preimages on the boundary sphereB , and consider f : B Rk. Prove that the number of preimages of z, countedwith our usual orientation convention, equals the winding number W(f,z).

Proof. To begin we appeal to the stack of records theorem (Exercise 7, Chapter 1,

Section 5, and stated in the appendix below). This tells us that f1

(z) is a finiteset {x1, . . . , xn} so that we can circumscribe disjoint balls Bi around each xi. Sincef1(z) is disjoint from B , we may also shrink these balls as necessary so thatBi B = and so that each Bi is sufficiently small in the sense of Exercise 1.

Let fi = f|Bi . Then, note that by Exercise 1 the number of preimage points,counted with our usual orientation convention, equals

ni=1 W(fi, z).

Let B = B \ni=1 Bi and consider the map u : B

Sk1, x f(x)z|f(x)z| . This

extends to all of B, since |f(x) z| = 0 on B. Thus, W(f|B , z) = deg(u) = 0,with the second inequality due to the proposition on page 110 of [1] (see appendix).However, since B = B

ni=1(Bi) (where the indicates that each Bi

inherits the opposite orientation it has when considered as a submanifold of B), weconclude that W(f|B , z) = W(f,z)

ni=1 W(fi, z). Therefore W(f,z) =n

i=1 W(fi, z).

Exercise 3. LetB be a closed ball inRk, and letf : Rk\Int(B) Y be any smoothmap defined outside the open ball Int(B). Show that if the restriction f : B Yis homotopic to a constant, then f extends to a smooth map defined on all of Rk

into Y.

Proof. For simplicity, assume B is centered at 0. Note that this allows us to writeevery non-zero point x B uniquely as x = ty for some y B and some t [0, 1].Let gt : B Y be a homotopy with g1 = f and g0 = constant. Now consider


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THE HOPF DEGREE THEOREM 3

the function F : Rk Y defined by

F(x) = f(x) if x Rk \ Int(B)

gt(y) if x B and x = ty for some y B and some t [0, 1] .

F is well-defined on Rk \ Int(B), since f and gt agree on the overlap (namely, thepoints in B where we have f = f = g1). F is also well-defined on B \ {0} sinceeach point in this set can be written uniquely as x = ty. Finally, F is well-definedat {0}, since g0 is a constant function. Furthermore, since the functions agree onthe overlap, F is continuous by the gluing lemma (as it is often called).

Now, using the standard trick of Exercise 1, Chapter 1, Section 6, we canperform a homotopy inside B, near B , in order to make F smooth on all ofRk

(note that it is already smooth on Rk \ B). Rather than do this explicitly, wewill speed things up and use a well-known approximation theorem which is not inGuillemin and Pollack (although this theorem is within the scope of the book). Youmay find this as Whitney Approximation on Manifolds in [2, Theorem 10.21] andstated in the appendix below. It allows up to homotope F rel Rk \ Int(B) to asmooth function. Then F is a smooth extension of f from all ofRk to Y.

3. The Special Case

Now that we have gotten some basic results out of the way, our next step will beto prove a special case of Hopfs theorem. The full theorem will derive from thisspecial case.

Special Case. Any smooth map f : Sl Sl having degree zero is homotopic to aconstant map.

Exercise 4. Check that the special case implies the following corollary.

Corollary. Any smooth map f : Sl Rl+1 \ {0} having winding number zero withrespect to the origin is homotopic to a constant.

Proof. By assumption, the degree of the map f /|f| is zero. So by the specialcase, f /|f| is homotopic to a constant. The map gt : S

l Rl+1 \ {0}, x

tf(x) + (1 t) f(x)|f(x)| , defines a homotopy with g0 = f /|f| and g1 = f. Since

homotopy is transitive, f is homotopic to a constant.

To prove the special case we will induct on dimension. Hopfs theorem, in thecase of 1-dimensional manifolds, is established as Exercise 9, Chapter 3, Section 3,and this will suffice as our base case. From here on, we will assume the special caseis true in dimension l = k 1. In order to complete the proof we need the followingexercise.

Exercise 5. Letf : Rk Rk be a smooth map with 0 as a regular value. Supposethat f1(0) is finite and that the number of preimage points in f1(0) is zerowhen counted with the usual orientation convention. Assuming the special case indimension k 1, prove that there exists a mapping g : Rk Rk \ {0} such thatg = f outside a compact set.

Proof. Since f1(0) is finite we can choose a ball B centered at the origin withf1(0) Int(B). The number of preimages is zero when counted with the usualorientation convention, by assumption, so Exercise 2 implies the map f : B


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Rk \ {0} has winding number zero. We know B is diffeomorphic to Sk1, so f is

a map from Sk1 into Rk \ {0}.

We are assuming the special case is true in dimension k 1, so the corollaryapplies. Thus, f is homotopic to a constant. At this point we wish to replace fwith the map f|Rk\Int(B) : R

k \ Int(B) Rk \ {0} so that we can apply Exercise

3. Then f extends to a smooth map g : Rk Rk \ {0} with f = g outside thecompact set B.

Notice that f and g are homotopic via the linear homotopy tf + (1 t)g, andthat outside of B this is constant.

All of the difficulty in the following proof will be showing that we can homotopef to miss a single point a of Sk. Once we have done this, it is easy to see that f ishomotopic to a constant map, for Sk \ {a} is contractible.

Exercise 6. Establish the special case in dimension k.

Proof. Using Sards theorem, pick distinct regular values a and b for f. We use thestack of records theorem again to tell us that f1(a) = {a1, . . . , an} and f1(b) =

{b1, . . . , bm} are finite. Thus, we can find an open neighborhood U of a1 such thatU is diffeomorphic to Rk via a diffeomorphism : Rk U and such that bi / U forall i. Note that we are assuming k > 1, since the base case is already established.This allows us to apply the corollary to the isotopy lemma of Chapter 3, Section 6(stated in the appendix) to each ai f1(a) with i = 1 to move ai into U via adiffeomorphism which is isotopic to the identity and compactly supported. Since weare only looking at f up to homotopy, this move does not concern us. Therefore, wehave a neighborhood U off1(a) such that U is diffeomorphic to Rk and b / f(U).

Let : Sk \ {b} Rk be a diffeomorphism such that (a) = 0. Then f is a smooth map from Rk into Rk. Since a is a regular value of f, 0 is a regularvalue of f , and since f1(a) is finite so is ( f )1(0). Since deg(f) = 0,

by assumption, the number of preimages of a is zero when counted with our usualorientation convention (as this is how we calculate degree when the image point isa regular value). It follows that the number of preimage points in ( f )1(0)is zero when counted with the usual orientation convention. Thus, we can applyExercise 5 to find a map h : Rk Rk \ {0} such that h = f outside acompact set B and h is homotopic to f on all ofRk. Note then, that onU, f is homotopic to 1 h 1, which is a map from U into Sk \ {a, b} since(1)1

{a}

=

{a}

= {0} (h 1)(U). Also note that on U \ 1(B) we

have f = 1 h 1, since h = f outside of B. This gives us a smoothmap g : Sk Sk \ {a} that is homotopic to f (take g = f on Sk \ 1(B) andg = 1 h 1 on 1(B)).

Since Sk\{a} is diffeomorphic to Rk, it is contractible. Therefore, g is homotopic

to a constant map, and it follows that f is as well.

4. The Hopf Degree Theorem

With the special case in hand we may now set our sights toward the goal, whichitself is essentially a special case of the following theorem.

Extension Theorem. LetW be a compact, connected, oriented k + 1 dimensionalmanifold with boundary, and let f : W Sk be a smooth map. Prove that fextends to a globally defined map F : W Sk, with F = f, if and only if thedegree of f is zero.


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Before proving the extension theorem we address the possibility of extendingboundary maps whose codomains are Rk+1.

Exercise 7. Let W be any compact manifold with boundary, and let f : W Rk+1 be any smooth map whatsoever. Prove that f may be extended to all of W.

Proof. By definition, W is a subset ofRN for some N, and since W is compact, itis a closed subset. Similarly, W is a closed subset ofRN. Since f is a smooth mapdefined on a closed subset ofRN it may be locally extended to a smooth map onopen sets. Since W is compact and boundaryless, we can use the -neighborhoodtheorem to extend f to a map F defined on a neighborhood U of W in RN.

Choose any smooth function that is one on W and zero outside some compactsubset of U. Then F can be defined on all ofRN, with the stipulation that it iszero outside U. This is an extension of f to all of RN since on W we haveF = 1 F = f. Then, F|W : W R

k+1 is an extension of f to all of W.

Exercise 8. Prove the Extension Theorem.

Proof. The sufficient direction follows from an application of the proposition onpage 110 of [1].

For the necessary direction we begin by extending f to a map F : W Rk+1

using Exercise 7. By the transversality extension theorem ([1, pg. 72], see appendix)we may assume 0 to be a regular value of F. For this, take Z = {0}, Y = Rk+1,C = W, and X = W in the theorem and replace the f in the theorem with ourF. The set F1(0) is finite (stack of records theorem), so we may use the corollaryto the isotopy lemma to move F1(0) so that F1(0) Int(B) where B is a closedball contained in Int(W).

Consider the map F = F|B : B Rk+1 \ {0}. Notice that F/|F| extends

from W to W

= W \ Int(B), since we moved F1

(0) inside Int(B). Thus,W(F|W , 0) = deg(F/|F|) = 0. We also know W(F|W, 0) = W(f, 0) = deg(f) =0, by assumption. Since W = W(B ), we have W(F|W , 0) = W(F|W, 0)W(F, 0) which implies W(F, 0) = 0.

By the corollary to the special case, F is homotopic to a constant. Therefore,it follows from Exercise 3 that F|W extends to a map F : W R

k+1 \ {0}. ThenF/|F| : W Sk is the global extension of f we were looking for.

Finally we have made it to the Hopf Degree Theorem. As important as it is, itis an almost trivial corollary of the extension theorem.

Exercise 9. Conclude:

The Hopf Degree Theorem. Two maps of a compact, connected, oriented k-manifold X into Sk are homotopic if and only if they have the same degree.

Proof. Let f0 and f1 be any two maps from X to Sk and let W = X I. Define

a map f : W Sk by setting f = f0 on X {0} and f = f1 on X {1}. Bythe extension theorem, f extends to a map on all of W if and only if deg(f) = 0.By definition, such an extension would be a homotopy of f0 and f1. Also, sinceW = (X {1})

(X {0})

it follows that deg(f) = deg(f1) deg(f0).

Thus, the extension theorem translates to: f0 and f1 are homotopic if and only ifdeg(f0) = deg(f1).


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5. Consequences

Now that any doubt as to the validity of Hopfs theorem has been eliminated, wemay explore some of its applications, the first of which is accessible to anyone whohas read far enough in [1] to understand the ideas presented above. This theoremis presented on page 146 of [1].

Existence of Nonvanishing Vector Fields. A compact, connected, orientedmanifold X possesses a nonvanishing vector field if and only if its Euler character-istic is zero.

Proof. This can be proved using hints given in the book.

An obvious application, which strays a bit from the subject of differential topol-ogy (into the realm of algebraic topology), involves homotopy groups of spheres.These are, in general, notoriously difficult to compute, but Hopfs theorem gives us

an easy way to do it in one very special case.The kth homotopy group of Sk. k(S

k) = Z, generated by the identity map.

Proof. Follows from Hopfs theorem by letting X = Sk and fixing basepoints.

With a little knowledge of homotopy theory, this corollary can be used to providean alternate proof (as opposed to the one on page 65 in [1]) that the unit k-diskdoes not retract onto its boundary which, in turn, can be used to prove the Brouwerfixed-point theorem.

Although the Hopf Degree Theorem is useful for applications, it is also interestingin its own right. You can examine the cases in dimensions one and two and you mayfind, in fact, that these low dimensional cases are intuitively true. But as we haveseen above, to actually prove it requires quite a bit of work, much like the Jordan-

Brouwer separation theorem. Of course, with work comes insight, so perhaps it isnot so terrible that these obvious statements are not so obvious afterall.

6. Appendix

The following statements were used during the course of the above proof, andwere added here to cater to the needs of the lazy reader who does not feel likeopening a copy of Guillemin and Pollack (or more specifically, to cater to the needsof the author who will be such a lazy reader if he decides to revisit this paper atsome point in the future). Proofs of these statements can be found in [1], unlessotherwise stated.

Linear Isotopy Lemma. Suppose that E is a linear isomorphism ofRk that pre-serves orientation. Then there exists a homotopy Et consisting of linear isomor-

phisms, such that E0 = E and E1 is the identity. If E reverses orientation, thenthere exists such a homotopy with E1 equal to the reflection map E1(x1, . . . , xk) =(x1, x2, . . . , xk).

Proof. To be provided by the reader. See Exercise 1, Chapter 3, Section 4 of [1] forassistance.

Stack of Records Theorem (Exercise 7, Chapter 1, Section 4). Suppose that yis a regular value of f : X Y, where X is compact and has the same dimensionas Y. Thenf1(y) is a finite set {x1, . . . , xN} and there exists a neighborhood U


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of y in Y such that f1(U) is a disjoint union V1 VN, where Vi is an openneighborhood of xi and f maps Vi diffeomorphically onto U.

Proof. To be provided by the reader.

Proposition (on page 110 of [1]). Suppose that f : X Y is a smooth mapof compact oriented manifolds having the same dimension and that X = W (Wcompact). If f can be extended to all of W, then deg(f) = 0.

Whitney Approximation on Manifolds. Let N and M be smooth manifolds,and let F : N M be a continuous map. Then F is homotopic to a smooth mapF : N M. If F is smooth on a closed subset A N, then the homotopy can betaken to be relative to A.

Proof. See theorem 10.21 in [2].

The 1-Dimensional Hopf Degree Theorem (Exercise 9, Chapter 3, Section 3).

Prove that two maps of the circle S1 into itself are homotopic if and only if theyhave the same degree.

Proof. To be provided by the reader.

Sards Theorem. If f : X Y is any smooth map of manifolds, then almostevery point in Y is a regular value of f. (i.e. The set of critical values of a smoothmap of manifolds f : X Y has measure zero).

Isotopy Lemma. Given any two points y and z in the connected manifold Y,there exists a diffeomorphism h : Y Y such that h(y) = z and h is isotopic tothe identity. Moreover, the isotopy may be taken to be compactly supported.

Corollary (to the Isotopy Lemma). Suppose that Y is a connected manifold ofdimension greater than 1, and let y

1, . . . , yn and z

1, . . . , zn be two sets of distinct

points in Y. Then there exists a diffeomorphism h : Y Y, isotopic to theidentity, with h(y1) = z1, . . . , h(yn) = zn. Moreover, the isotopy may be taken tobe compactly supported.

-Neighborhood Theorem. For a compact boundaryless manifold Y inRM anda positive number , letY be the open set of points inRM with distance less than fromY. If is sufficiently small, then each point w Y possesses a unique closestpoint in Y, denoted (w). Moreover, the map : Y Y is a submersion. WhenY is not compact, there still exists a submersion : Y Y that is the identityon Y, but now must be allowed to be a smooth positive function on Y, and Y isdefined as {w RM | |w y| < (y) for some y Y}.

Transversality Extension Theorem. Suppose that Z is a closed submanifold of

Y, both boundaryless, and C is a closed subset of X. Let f : X Y be a smoothmap with f Z on C and f Z on C X. Then there exists a smooth mapg : X Y homotopic to f, such that g Z, g Z, and on a neighborhood of Cwe have g = f.

References

[1] V. Guillemin, A. Pollack, Differential Topology, Prentice-Hall, Englewood Cliffs, NJ, 1974.[2] J. Lee, Introduction to Smooth Manifolds, Springer, New York, NY, 2002.[3] A. Hatcher, Algebraic Topology, Cambridge University Press, New York, NY, 2001.

hopf degree theorem

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