From calculus to cohomology: de Rham cohomology and characteristic classes

278 Pages·1997·20.372 MB·English

by  Ib Henning Madsen, Jørgen Tornehave

From Calculus to Cohomology

.
de Rham cohomology and characteristic classes
Ib Madsen and J~rgen Tornehave
University q{Aarhus
,,~~I CAMBRIDGE
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UNIVERSITY PRESS
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Contents
.:....
',)
'­
.
. ~)<,
fJ Preface .vii

-.

,~ Chapter 1
Introduction . . . . . . . . . 1

f"J J
Chapter 2
The Alternating Algebra. . 7

Chapter 3
de Rham Cohomology. . 15

Chapter 4
Chain Complexes and their Cohomology 25

Chapter 5
The Mayer-Vietoris Sequence . . . . . 33

Chapter 6
Homotopy . . . . . . . . . . . . . . . . . 39

Chapter 7
Applications of de Rham Cohomology 47

Chapter 8
Smooth Manifolds . . . . . . . . . . . . 57

Chapter 9
Differential Forms on Smoth Manifolds . 65

Chapter 10
Integration on Manifolds 83

Chapter II
Degree, Linking Numbers and Index of Vector Fields. 97 'j

I

Chapter 12
The Poincare-Hopf Theorem . . . . . II 3

,I
Chapter 13
Poincare Duality . . . . . . . . . . . . 127 j.
j}

Chapter 14
The Complex Projective Space ern . 139 ~
Chapter 15
Fiber Bundles and Vector Bundles. . 147

Chapter 16
Operations on Vector Bundles and their Sections 157

Chapter 17
Connections and Curvature . . . . . . . . . . . . . 167

Chapter 18
Characteristic Classes of Complex Vector Bundles 181

Chapter 19
The Euler Class. . . . . . . . . . . . . . . . . . . . . 193

Chapter 20
Cohomology of Projective and Grassmannian Bundles 199

Chapter 21
Thorn Isomorphism and the General Gauss-Bonnet Formula . 211

Appendix A Smooth Partition of Unity. . . . .
. 221

Appendix B Invariant Polynomials . . . . . . .
. 227

Appendix C Proof of Lemmas 12.12 and 12.13
. 233

Appendix D Exercises
. 243

References . 281

Index ... . 283
VB
PREFACE
This text offers a self-contained exposition of the cohomology of differential
forms, de Rham cohomology, and of its application to characteristic classes
defined in terms of the curvature tensor. The only formal prerequisites are
knowledge of standard calculus and linear algebra, but for the later part of the
book some prior knowledge of the geometry of surfaces, Gaussian curvature, will
not hurt the reader.
The first seven chapters present the cohomology of open sets in Euclidean spaces
• and give the standard applications usually covered in a first course in algebraic
topology, such as Brouwer's fixed point theorem, the topological invariance of
domains and the Jordan-Brouwer separation theorem. The next four chapters
extend the definition of cohomology to smooth manifolds, present Stokes' the­
orem and give a treatment of degree and index of vector fields, from both the
cohomological and geometric point of view.
The last ten chapters give the more advanced part of cohomology: the
Poincare-Hopf theorem, Poincare duality, Chern classes, the Euler class, and
finally the general Gauss-Bonnet formula. As a novel point we prove the so
called splitting principles for both complex and real oriented vector bundles.
The text grew out of numerous versions of lecture notes for the beginning course
in topology at Aarhus University. The inspiration to use de Rham cohomology as
a first introduction to topology comes in part from a course given by G. Segal at
Oxford many years ago, and the first few chapters owe a lot to his presentation
of the subject. It is our hope that the text can also serve as an introduction to the
modern theory of smooth four-manifolds and gauge theory.
The text has been used for third and fourth year students with no prior exposure
to the concepts of homology or algebraic topology. We have striven to present
all arguments and constructions in detail. Finally we sincerely thank the many
students who have been subjected to earlier versions of this book. Their comments
have substantially changed the presentation in many places.
Aarhus, January 1996
1. INTRODUCTION
It is well-known that a continuous real function, that is defined on an open set of
R has a primitive function. How about multivariable functions? For the sake of
simplicity we restrict ourselves to smooth (or Coo-) functions, i.e. functions that
have continuous partial derivatives of all orders.
We begin with functions of two variables. Let f: U ---; R2 be a smooth function
.defined on an open set of R2 .
Question 1.1 Is there a smooth function F: U ---; R, such that
of of
(1) -;- = hand -;- = h, where f = (h, h)?
VXI VX2
Since
02F 02F
aX2aXI aXl aX2
we must have
ah ah
(2)
=
aX2 aXI'
The correct question is therefore whether F exists, assuming f = (h, h) satisfies
(2). Is condition (2) also sufficient?
Example 1.2 Consider the function f: R2 ---; R2 given by
f(xI,x2)= ( xi-X+2x rxiX+xl~) .
It is easy to show that (2) is satisfied. However, there is no function F: R2 -{0} ---;
R that satisfies (1). Assume there were; then
(2Jr d
Jo dBF(cos B, sin B)dB = F(l, 0) - F(l, 0) = o.
On the other hand the chain rule gives
of
d . dF .
dBF(cosB, smB) = dx . (-smB) + ay . cosB
= - h(cosB, sinB) . sinB + h(cosB, sin B)· cosB = 1.
This contradiction can only be explained by the non-existence of F.
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2 1. INTRODUCTION
Definition 1.3 A subset X c IRn is said to be star-shaped with respect to the Note that rot 0 grad = I
point Xo E X if the line segment {t:ro + (1- t)xlt E [0, I]} is contained in X
for all x E X.
Theorem 1.4 Let U C 1~2 be an open star-shaped set. For any smooth function
Since both rot and grad
(!I, h): U ---+ 1R 2 that satisfies (2), Question 1.1 has a solution.
Therefore we can cons
casets + Im(grad) w
2 Do
Proof. For the sake of simplicity we assume that Xo = 0 E 1R . Consider the

function F : U ---+ IR, (3)

i
1
F(Xl,X2) = [Xl!I(txl,tx2) +x2h(txl,t:rz)]dt. Both Ker(rot) and Im(g
able that the quotient sI
Then one has We can now reformulat
1
~a(F: rl, :r2) = 1 [it. (tXl' tX2) + tXl ~a(!It Xl, tX2) + tX2a~(t!Xl,z tX]2) dt (4) H 1(U)
U.rl 0 UXI UXI
On the other hand, Exarr
and
see that HI (1~2 - {O}) i
d a!I a!I dimension of H1(U) is
-t!I(tXI, tX2) = !I(tXl, tX2) + tXl ~ (tXl, tX2) + tX2~(tXI, tX2).
dt UXI UX2 In analogy with (3) we
Substituting this result into the formula, we get
(5)
rl
[d
aa:Frl (Xl,X2) = Jo dtt!I(txl,tx2)+tx2 (aa!Xzl (tXI,txZ)- aaX!2I (txl, tx2) )] dt This definition works fo
= [t!I (txl, tX2)]Z=O = !I(Xl, X2).
Analogously, g~ = !z(Xl, X2).
Theorem 1.5 An open i
"Example 1.2 and Theorem 1.4 suggest that the answer to Question 1.1 depends
on the "shape" or "topology" of U. Instead of searching for further examples or Proof. Assume that gra
counterexamples of sets U and functions f, we define an invariant of U, which tells a neighborhood V (xo) ,
us whether or not the question has an affirmative answer (for all j), assuming then every locally const:
the necessary condition (2).
Given the open set U <;;;; 1R2, let Coo (U, IRk) denote the set of smooth functions {x
¢: U ---+ IRk. This is a vector space. If k = 2 one may consider ¢: U ---+ Rk as is closed because f is (
a vector field on U by plotting ¢(u) from the point u. We define the gradient
it is equal to U, and H(
and the rotation
exists a smooth, surject
grad: COO(U, IR) ---+ Coo (U, 1R2), rot: Coo (U, 1R2) ---+ COO(U, IR) constant, so grad(J) = (
by The reader may easily e
a¢ a¢) a¢l a¢2 <
is precisely the number
grad(¢) = ( aXl' aX2 ' rot(¢1,¢2) = aX2 - aXl
L
1. INTRODUCTION 3
Note that rot grad = O. Hence the kernel of rot contains the image of grad,
0
Ker(rot) = Kernel of rot

Im(grad) = Image of grad

Since both rot and grad are linear operators, Im(grad) is a subspace of Ker(rot).
Therefore we can consider the quotient vector space, i.e. the vector space of
cosets 0: + Im(grad) where 0: E Ker(rot):
(3) HI(U) = Ker(rot)jlm(grad).
Both Ker(rot) and Im(grad) are infinite-dimensional vector spaces. It is remark­

able that the quotient space HI (U) is usually finite-dimensional.

We can now reformulate Theorem 1.4 as

(4) HI(U) = 0 whenever U ~ 1R 2 is star-shaped.
On the other hand, Example 1.2 tells us that HI (R2 - {O}) -=I O. Later on we shall
see that HI (1R2- {O}) is I-dimensional, and that HI (1R2 - U7=1 {Xi}) ~ IRk. The
dimension of HI (U) is the number of "holes" in U.
In analogy with (3) we introduce
(5) HO(U) = Ker(grad).
This definition works for open sets U of I~k with k 2: 1, when we define
(:f of )
grad(J) =
Xl 1"" Oxn '
Theorem 1.5 An open set U ~ Rk is connected if and only if HO(U) = R
Proof. Assume that grad(J) = O. Then f is locally constant: each Xo E U has
a neighborhood V(xo) with f(x) = f(xo) when X E V(xo). If U is connected,
then every locally constant function is constant. Indeed, for Xo E U the set
{X E Ulf(x) = f(xo)} = f-I(J(xo)).
is closed because f is continuous, and open since f is locally constant. Hence
it is equal to U, and HO(U) = R. Conversely, if U is not connected, then there
exists a smooth, surjective function f: U --+ {O, I}. Such a function is locally
constant, so grad(J) = O. It follows that dimHO(U) > 1. 0
The reader may easily extend the proof of Theorem 1.5 to show that dimHO(U)
is precisely the number of connected components of U.
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4 1. INTRODUCTION
We next consider functions of three variables. Let U c;;: R3 be an open set. A real Example 1.7 Let S = {(:l
function on U has three partial derivatives and (2) is replaced by three equations. in the (Xl, X2 )-plane. Cor
We introduce the notation
f(Xl' X2, X3)
grad: Coo(U, R) ---+ Coo(U, R3 )
-2XlX3
rot: Coo (U, R3) ---+ Coo (U, R3) (
div: Coo(U, R3) ---+ Coo(U, R) X~ + (xi + X~ ­
for the linear operators defined by on the open set U = R3 ­
af af af ) One finds that rot(J) =
grad (J) = ( -a' -a' -a integration along a curve "I
Xl X2 X3
we shall show that [I) i= (
rot (h, 12, h) = (a13 _ a12 , ah _ a13 , a12 _ ah )
aX2 aX3 aX3 aXl aX2 aX2
. ah ah ah I'(t)=(,
dlv(h,h,h)=-a +-a +-a.
X2 X3
Xl Assume grad(F) = f as
Note that rot 0 grad = 0 and div 0 rot = O. We define HO(U) and set Hl(U) :ftF(ry(t)) in two ways. 0
as in Equations (3) and (5) and
(6) H2 (U) = Ker(div)jlm(rot). j 1r-E d F(I'(t))dt =
dt

-1r+E
Theorem 1.6 For an open star-shaped set in R3 we have that HO (U) R

and on the other hand the
Hl(U) = 0 and H 2(U) = O.

d

dtF(ry(t)) = fl(ry(t
Proof. The values of HO(U) and Hl(U) are obtained as above, so we shall
restrict ourselves to showing that H 2(U) = O. It is convenient to assume that U = sin2t -j
is star-shaped with respect to O. Consider a function F : U ---+ R3 with div F = 0,
Therefore the integral also,
and define G : U ---+ R3 by
1
1
Example 1.8 Let U be an
G(x) = (F(tx) x tx)dt
smooth vector field). Recal
where x denotes the cross product, r: [a, b] ---+ U is defined by
h
el Xl

=1
(h, 12, h) x (Xl, X2, X3) e2 12 X2
A.,(
13 X3
e3
= (hx3- h X2, hXl - h X3, h X2- hXl). where (,) denotes the stan
Straightforward calculations give 4>,,( b), then the energy is ,
d
rot(F(tx) x tx) = dt(t2 F(tx)). (X
Hence
rotG(x) = 11 -d( t2F(tx))dt = F(x). o by the chain rule; compare
°
dt
If U c;;: R3 is not star-shaped both Hl(U) and H 2(U) may be non-zero.
i,...
1. INTRODUCTION 5
Example 1.7 Let S = {(Xl, X2, X3) E 1R31xi + X§ = 1, X3 = o} be the unit circle
in the (Xl, X2 )-plane. Consider the function
f(:1:I' X2, X3)
_ (-2XIX3 -2X2X3 xi + x§ - 1 )
- x32 +(x,.2l +x22 -1) 2' X32 +(X2I +x22 -1) 2' .x23 +(x2l +x22 -1) 2
on the open set U = R3 - S.

One finds that rot(f) = 0. Hence f defines an element [1] E HI(U). By

integration along a curve, in U, which is linked to S (as two links in a chain),

we shall show that [1] i- 0. The curve in question is

,(t) = (VI +cos t,O, sin t ), -Jr:S t :s Jr.
Assume grad(F) = f as a function on U. We can determine the integral of
fit
F(r(t)) in two ways. On the one hand we have
j 7r-€ d ° °
-dF(r(t))dt = F(r(Jr - f)) - F(r(-Jr + f)) -+ for f -+
t
-7r+€
and on the other hand the chain rule gives
,~(t) ,~(t) ,~(t)
:tF(r(t)) = !t(,(t)) . + J2(r(t))· + h(,(t)).
°
= sin2t + +cos 2t = 1.
Therefore the integral also converges to 2Jr, which is a contradiction.
Example 1.8 Let U be an open set in IRk and X: U -+ IRk a smooth function (a
smooth vector field). Recall that the energy A')'(X), of X along a smooth curve
,: [a, b] -+ U is defined by the integral
l
b
A')'(X) = (X o,(t) , ,'(t))dt
where ( ,) denotes the standard inner product. If X = grad (~) and ~,( a) =
~,( b), then the energy is zero, since
d
(X 0 ,(t) , ,'(t)) = dt ~(r(t))
by the chain rule; compare Example 1.2.
7
2. THE ALTERNATING ALGEBRA
Let V be a vector space over I~. A map
f:VxVx···xV-d~
, .... J
k times
is called k-linear (or multilinear), if f is linear in each factor.
Definition 2.1 A k-linear map w: V k ---+ IR is said to be alternating if
w(6, ... ,~k) = 0 whenever ~i = ~j for some pair i I- j. The vector space
of alternating, k-linear maps is denoted by Altk (V).
We immediately note that Altk(V) = 0 if k > dim V. Indeed, let el, ... , en be a
basis of V, and let W E Altk(V). Using multilinearity,
w(6"",~k)
=W(LAi,lei""'LAi,kei) = LAJw(eh,···,ejk)
with AJ = Ail,l Ajk,k' Since k > n, there must be at least one repetition among
the elements eh, , ejk' Hence w(cil' ... , ejk) = O.
The symmetric group of permutations of the set {I, ... ,k} is denoted by S(k).
We remind the reader that any permutation can be written as a composition of
transpositions. The transposition that interchanges i and j will be denoted by (i, j).
Furthemiore, and this fact will be used below, any permutation can be written as a
composition of transpositions of the type (i, i+1), (i, i+1)o(i+1, i+2)o(i, i+1) =
(i, i + 2) and so forth. The sign of a permutation:
(1) sign: S(k) ---+ {±1},
is a homomorphism, sign(o- 0 T) = sign(o-) 0 sign(T), which maps every transpo­
sition to -1. Thus the sign of 0- E S(k) is -1 precisely if 0- decomposes into a
product consisting of an odd number of transpositions.
Lemma 2.2 IfW E Altk(V) and 0- E S(k), then
W(~a(l)"" ,~a(k)) = sign(o-)w(6,··· ,~k)'
Proof. It is sufficient to prove the formula when 0- = (i, j). Let
Wi,j (~, ~') = w(6, .. · ,~, ... ,(, ... , ~k),
e
with ~ and occurring at positions i and j respectively. The remaining ~v E V
are arbitrary but fixed vectors. From the definition it follows that Wi,j E Alt2 (V).
Hence Wi,j (~i + ~j, ~i + ~j) = O. Bilinearity yields that Wi,j (~i, ~j) + Wi,j (~j, ~i) =
O. 0
-:.
.. £"i->~ .,... . " <IS
1
.~ ..
~ '9,
~ ~
!7.· ",,,_-,,c·;;..p,=~_
. . r rap
L
sTT] .•
8 2. THE ALTERNATING ALGEBRA
Example 2.3 Let V = IRk and ~i = (~il"",~ik). The function w(6"",~k) = Lemma 2.7 A k-linear m
det( (~ij)) is alternating, by the calculational rules for determinants. with ~i = ~i+l for some:
We want to define the exterior product Proof. 5 (k) is generated
of Lemma 2.2,
/\: AltP(V) x Altq(V) --+ Altp+q(V).
W(6"",~i,~:
When p = q = 1 it is given by (Wl/\ W2) = wl(6)W2(6) - w2(6)Wl(6).
Hence Lemma 2.2 holds ti
Definition 2.4 A (p, q)-shuffle fJ is a permutation of {I, ... ,p + q} satisfying
It is clear from the defini
fJ(l) < ... < fJ(p) and fJ(p + 1) < ... < fJ(p + q).
(WI'
The set of all such permutations is denoted by 5(p, q). Since a (p, q)-shuffle (AWl
is uniquely determined by the set {fJ(l), ... ,fJ(p)}, the cardinality of 5(p,q) is
WI/\
(p;q).
for wI, W~ E AltP(V) and
Definition 2.5 (Exterior product) For WI E AltP(V) and W2 E Altq(V), we define
(WI /\ W2)(6,···, ~p+q)
Lemma 2.8 IfWI E AltPC
L
= sign(fJ)wl (~a(I)'"'' ~a(p») . W2(~a(p+l)'"'' ~a(p+q))'
Proof. Let T E S(p + q)
aES(p,q)
T(l) =
It is obvious that WI /\ W2 is a (p + q)-linear map, but moreover:
T(q+ 1) =
Lemma 2.6 If WI E AltP(V) and W2 E Altq(V) then WI /\ W2 E Altp+q(V). We have sign(T) = (-1)P'i
S(
Proof. We first show that (Wl/\W2)(6,6, ... ,~p+q) = 0 when ~1 = 6. We let
Note that
(i) 512 = {fJ E 5(p, q) IfJ(l) = 1, fJ(p + 1) = 2} (~aT(1
(ii) 521 = {fJ E 5(p, q) IfJ(l) = 2, fJ(p + 1) = I} W2
(~aT(q+l)"
(iii) 50 = 5(p, q) - (512 U 521)' WI
Hence
If fJ E 50 then either WI (~a(I)' ... , ~a(p») or W2(~a(p+l)' ... ,~a(p+q)) is zero, since
~a(l) = ~a(2) or ~a(p+l) = ~a(p+2)' Left composition with the transposition W2 /\ wI(6, ... ,~p
L
T = (1,2) is a bijection 512 --+ 521. We therefore have sign(fJ:
(WI /\ W2) (6,6, ... ,~p+q) = aES(q,p)
L
L sign(a~
sign(fJ )Wl (~a(I)' ... , ~a(p»)W2 (~a(p+l)' ... ,~a(p+q))
aES(p,q)
aE S12 L
- L
sign(fJ)wl (~Ta(I)" .. '~Ta(p)) . W2 (~Ta(p+l)" .. ~Ta(p+q)). = (-l)Pq
aES12 aES(p,q)
Since fJ(l) = 1 and fJ(P+ 1) = 2, while TfJ(l) = 2 and TfJ(P+ 1) = 1, we see that
= (-l)PqWI/\ W2
TfJ( i) = rJ(i) whenever i =j:. 1, p + 1. But 6 = ~2 so the terms in the two sums

=
cancel. The case ~i ~i+l is similar. Now WI /\ W2 will be alternating according

to Lemma 2.7 below. D