analysis on-manifolds notes
TRANSCRIPT
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Analysis on Manifolds
Based on the lectures of Professor
Leonid Polterovich
Daniel Rosen
Semester B, 2007
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Warning: This text has not been thoroughly checked and might contain
inconsistencies and even mistakes
Contents
1 Smooth manifolds and maps 3
1.1 Topological manifolds . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Smooth manifolds . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.3 Smooth maps . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2 The tangent space and tangent bundle 9
2.1 The tangent space . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 The tangent bundle . . . . . . . . . . . . . . . . . . . . . . . . 11
2.3 Submersions and immersions . . . . . . . . . . . . . . . . . . . 13
3 Boundary and orientation 17
3.1 Manifolds with boundary . . . . . . . . . . . . . . . . . . . . . 17
3.2 Orientation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.3 The Partition of Unity . . . . . . . . . . . . . . . . . . . . . . 22
3.4 Riemannian manifolds . . . . . . . . . . . . . . . . . . . . . . 24
4 Homotopy and degree of a map 26
4.1 Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.2 The Degree Theorem (I) . . . . . . . . . . . . . . . . . . . . . 28
4.3 Euler characteristic and vector fields . . . . . . . . . . . . . . 33
5 Exterior differential calculus 40
5.1 Exterior forms on linear spaces . . . . . . . . . . . . . . . . . 40
5.2 Differential forms in Rn . . . . . . . . . . . . . . . . . . . . . . 445.3 The exterior derivative . . . . . . . . . . . . . . . . . . . . . . 45
5.4 Differential forms on manifolds . . . . . . . . . . . . . . . . . 48
6 Integration on manifolds 50
6.1 Integrals of differential forms . . . . . . . . . . . . . . . . . . . 50
6.2 Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 52
6.3 The Degree Theorem (II) . . . . . . . . . . . . . . . . . . . . . 57
6.4 Linking number and the Gauss integral . . . . . . . . . . . . . 60
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7 The Lie derivative 64
7.1 The Lie derivative . . . . . . . . . . . . . . . . . . . . . . . . . 64
7.2 Application: Hamiltonian mechanics . . . . . . . . . . . . . . 65
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1 Smooth manifolds and maps
1.1 Topological manifolds
Definition 1.1.1. A topological manifold of dimension m is a topological
spaceM in which every x M has a neighborhood which is homeomorphic toan open set in Rm. We use the notation Mm for an m-dimensional manifold.
Remark. The dimension of a topological manifold is well-defined. This
follows from the Invariance of Domain Theorem, by which open sets in Rnand Rm are homeomorphic only for n = m.
Example 1.1.2. The n-sphere: Sn = {x Rn+1 : |x| = 1}.Denote by P the north and south poles, and set U = Sn \ {P}. Definethe Stereographic Projection : U Rn as follows:
Figure 1: Stereographic Projection
Claim. The stereographic projection is a homeomorphism
Since {U} is an open cover of Sn, this defines Sn as a topological mani-fold.
Example 1.1.3 (Product manifolds). If Mm and Nn are topological mani-
folds, thenMN , with the product topology, can be defined as a topologicalmanifold of dimension m+ n in an obvious manner.
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Example 1.1.4. The n-dimensional torus Tn. We give two equivalent defi-nitions:
(i) Tn = S1 . . . S1 n times
, with the product topology.
(ii) Tn = Rn/Zn. Here we use the quotient topology on Tn and define thetopological manifold structure using the natural projection,
pi : Rn Tn, x 7 [x]
This construction is explained in more detail later.
Example 1.1.5. The n-dimensional real projective space RP n. Again, wegive two definitions:
(i) RP n = {All lines in Rn+1 passing through the origin}. Define a metricon RP n by
d(l1, l2) = ](l1, l2) [0, pi)where the angle is measured by the standard Euclidean inner product,
and use the metric topology.
(ii) Define an equivalence relation on Sn: x y x = y, and setRP n = Sn/ . Again, use the natural projection to define a topologicalmanifold structure.
1.2 Smooth manifolds
Let M be a topological manifold.
Definition 1.2.1. A chart on M is a pair (U, ), where U M and(U) Rn are open and : U (U) is a homeomorphism.On a chart (U, ), consider the functions xi defined by (a) =
(x1(a), . . . , xn(a)
).
These functions are called local coordinates on U .
Definition 1.2.2. Let (U, ) and (V, ) be two charts with U V 6= .Their transition function is 1 : (U V ) (U V ), which is ahomeomorphism between open sets in Rn.
Definition 1.2.3. Let U Rn be an open set. A function f : U Rm iscalled smooth if its partial derivatives of all orders exist and are continuous.
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Definition 1.2.4. A smooth atlas on M is a collection {(U, )} of chartswhich satisfies:
1. M =
U
2. Whenever U U 6= , the transition function 1 is smooth.Example 1.2.5. The 1-sphere:
Claim. {(U, )} is a smooth atlas on S1.Proof. Straightforward computation shows that for any x 6= P we have+(x)(x) = 1. Now, +(U+ U) = (U+ U) = R \ {0}. Thereforethe transition function is
R \ {0} R \ {0} , t 7 1t
which is smooth.
Example 1.2.6. The torus Tn : We use the natural projection pi : Rn Zn.The problem is that pi is not injective. To get around that, we take open
cubes Q of size length 12, on which pi is injective. Denote
Q = piQ: Q Tn
Then {Q(Q) : Q - open cube of side length 12} is an open cover of Tn. Fi-nally, take the atlas {(Q(Q), 1Q )}. The transition functions are simplytranslations, so this is a smooth atlas.
Definition 1.2.7. Two smooth atlases on M are said to be compatible if
their union is still a smooth atlas.
Definition 1.2.8. A smooth structure onM is an equivalence class of smooth
atlases. A manifold M together with a smooth structure is called a smooth
manifold. From now on, unless otherwise noted, the word manifold will mean
a smooth manifold.
Remark. It is easy to see that an arbitrary union of compatible atlases is
still a smooth atlas compatible with all atlases from the union. Therefore,
any smooth structure contains a maximal atlas.
Definition 1.2.9. Let Mm be a smooth manifold. A subset N M iscalled a submanifold of dimension n if for any x N there is a chart (U, )on M such that U N = 1(Rn). Here we consider Rn as a subset ofRm = Rn Rmn, namely Rn {0} . Of course, n m.
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Remark. The collection {(U N, UN)} forms a smooth atlas on N .
Thus any submanifold is in itself a manifold.
Note that the submanifold condition is local, as opposed to that of a man-
ifold. That is because the global patching together of charts is guaranteed
by the ambient manifold.
Theorem 1.2.10. Let M Rn be a subset such that for any x M thereexists an open set U and smooth functions f1, . . . , fk : U R such thatMUis the common zero set of the functions fi , i.e.
M U = {f1 = . . . = fk = 0}
and
rank
(fixj
(y)
)= k y M U
Then M is a manifold of dimension m := n k.Stated slightly differently, F = (f1, . . . , fk) : U Rk is smooth and for ally M U , rankDyF = k.Proof. Denote elements in Rn by (x, y), x Rm, y Rk. Let (a, b) M ,and U, F as given. WLOG, assume that the last k columns in DF are
linearly independent. Then by the Implicit Function Theorem there exist
neighborhoods a V Rm , b W Rk and a unique f : V W suchthat for any x V , y W we have F (x, y) = 0 y = f(x). ThenM U = {(x, f(x)) : x V }. Define
: M U V W Rn , (x, y) = (x, f(x) y)
Then is a diffeomorphism. For any p M we can find such (Up, p), andthen {(Up, p)} is an smooth atlas on M .
Example 1.2.11. The n-sphere. Define
F : Rn+1 R , F (x) = |x|2 1
Then Sn = {F = 0}. Now,
DxF = xF = (2x1, . . . , 2xn+1) x Sn.xF 6= 0
In other words, rankF = 1 throughout Sn, proving that Sn is an n-dimensional manifold.
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Example 1.2.12. The special linear group,
SL(n) = {A Mn(R) : det(A) = 1}
Since Mn(R) Rn2 , we can use the previous theorem. Define:
F : Mn(R) R , F (A) = det(A) 1
then SL(n) = {F = 0}. Well use the following
Claim.d
dt
t=0
det(I + tB) = tr(B)
Let Eij be the standard basis for Mn(R). For A SL(n), we have
F (A+ tEij) = det(A) =1
det(I + tA1Eij) = det(I + tA1Eij)
So by the claim
F
xij(A) =
d
dt
t=0
F (A+ tEij) = tr(A1Eij)
Denote the entries of A1 by aij. Then
A1Eij =
0 a1i . . . 00 a2i . . . 0. . . . . . . . . . . . . . . . . .
0 ani . . . 0
jth column
tr(A1Eij) = aij
Since A 6= 0, there is at least one aij 6= 0 and so DF 6= 0. As before, thisimplies that SL(n) is a manifold of dimension n2 1.
1.3 Smooth maps
Let Mm, Nn be manifolds, and f : M N a continuous map.Definition 1.3.1. Let x M , and let (U, ) and (V, ) be charts on M andN , respectively, such that f(U) V . Denote f = f 1 : (U) (V ).The triplet (U, ) , (V, ) , f is called a local representation of f (around x).
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If (U0, 0) , (V0, 0) , f0 is another local representation of f , Then
f0 = 0 f 10 = 0 1 T1
f 1 f
10 T2
= T1 f T2
where T1 , T2 are (smooth) transition functions. In particular, the smoothness
of one local representation implies the smoothness of all others.
Definition 1.3.2. f is called smooth at x if there is a local representation of
f around x which is smooth. f is called smooth if it is smooth at all x M .Definition 1.3.3. A map f is called a diffeomorphism if it is invertible, and
both f and f1 are smooth.
Smooth maps naturally induces maps between smooth real-valued func-
tions.
Definition 1.3.4. The set of all real-valued function onM is denoted F(M).The set of smooth real-valued functions on M is denoted C(M).
Proposition 1.3.5. F(M) is a real algebra, i.e. it is a real linear spaceequipped with a bilinear multiplication - ()(x) = (x)(x). C(M) F(M) is a subalgebra.Definition 1.3.6. Let f : M N be a smooth map. It induces the pullbackmap:
f : C(N) C(M) , 7 fClaim. The pullback f is a homomorphism of algebras. This means that itpreserves addition and multiplication.
The pullback can be defined for general functions, as a homomorphism
F(N) F(M). This gives a different characterization of smooth maps.Excercise 1.3.7. A map f : M N is smooth iff f (C(N)) C(M)
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2 The tangent space and tangent bundle
2.1 The tangent space
The tangent space to a manifold at a given point is a linear approximation
of the manifold around that point. In Rn, it is defined by velocity vectors ofsmooth curves: A curve through x Rn is a smooth : (1, 1) Rn suchthat (0) = x. Its velocity vector at x is
(0) :=d
dt
t=0
(t)
The tangent space at x Rn is defined asTxRn = {(0) : - smooth curve through x}
It is easy to see that TxRn Rn.We want to extend the definition of a tangent space to an arbitrary man-
ifold. The difficulty is that we do not have the concept of a velocity vector.
We get around that in the following manner:
Definition 2.1.1. A smooth curve through x M is a smooth function : (1, 1) M such that (0) = x. We deal below with smooth curves,unless otherwise noted.
Note that if (U, ) is a chart around x M and is a curve through xthen : (1, 1) (U) is a curve through (x) Rn.Definition 2.1.2. Let x M , and let (U, ) be a chart around x. Twocurves 1, 2 through x are called equivalent, denoted 1 2, if (1)(0) =( 2)(0).
It can easily be shown that this definition does not depend on the choice
of chart.
Definition 2.1.3. The tangent space to M at x, denoted TxM , is the set of
equivalent classes of smooth curves through x.
The tangent space in Rn is by definition a linear space. The natural linearstructure on a tangent space to a manifold is defined by pulling back the
linear structure from Rn.
Claim. The map pi : TxM T(x)Rn , [] 7 ( )(0) is well defined and abijection.
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Definition 2.1.4. We define operations on TxM as follows:
[] = pi1( pi[]) [1] + [2] = pi1(pi[1] + pi[2])
Claim. These operations are well-defined, i.e. do not depend on the choice
of chart, and turn TxMn into an n-dimensional linear space.
The tangent space can be defined differently, in terms of derivations of
the algebra C(M).
Definition 2.1.5. A derivation of C(M) at x M is an R-linear functionalD C(M) which satisfies the Leibnitz rule:
D(fg) = D(f)g(x) + f(x)D(g)
The set of all derivations of C(M) at x is denoted Derx(C(M)
)Claim. Derx
(C(M)
)is a linear subspace of the dual space C(M).
Definition 2.1.6. For [] TxM , define
D Derx(C(M)
), D(f) =
d
dt
t=0
f
It is easy to see that D is well-defined, i.e. does not depend on choice of
representative. Clearly, it is indeed a derivation.
Excercise 2.1.7. The map TxM Derx(C(M)
), [] D is an isomor-
phism of linear spaces.
So, we can view the tangent space TxM as the space of derivation at x.
In particular, if x1, . . . , xn are local coordinates around a M , we can de-fine the following derivations through the local representation - the partial
derivatives:
if :=f
xi(a)
Since these are n derivations, and they are obviously linearly independent
(as ixj = ij), they form a basis for Dera
(C(M)
)= TaM .
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Example 2.1.8 (Zero sets in Rn). LetM Rn be a the zero set of a smoothfunction F : Rn R. We identify TxRn Rn, then
TxM TxRn = {(0) : - smooth curve through x in M}
Let be such a curve, then (t) M , and so F((t)) 0. Differentiatingwe get:
DxF((0)
)=(xF, (0)) = 0
So,
TxM = kerDxF = {xF}
More generally, if M is the zero set of F : Rn Rm, then
TxM = kerDxF
2.2 The tangent bundle
The tangent space is a local construction. Globalizing it, we define the
tangent bundle:
Definition 2.2.1. The tangent bundle to M is TM =xM
TxM . Elements
in TM are of the form (x, ) , x M , TxM .Example 2.2.2. Let U Rn be open. Then there is a natural isomorphismTxM Rn at each x U , so TU M Rn.
To define the concept of smooth vector fields on M , we need to define on
TM a smooth structure.
Definition 2.2.3. Given an atlas {(U, )} on M , define U =
xUTxM .
Since (U) Rn is open, as noted above,
T(U) (U) Rn R2n
Define:
: U T(U) , (x, ) 7((x), Dx()
)Thus each maps U onto an open subset of R2n. In order to make
{(U, )} a smooth atlas, we first need a topology on TM .Definition 2.2.4. X TM is open if for any , (X U) R2n is open.
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Note that since is already bijective, this definition automatically makes
it a homeomorphism.
Proposition 2.2.5.
This topology is well-defined. With this topology TM becomes a topological manifold. {(U, )} then becomes a smooth atlas on TM .The definition of the tangent space and bundle allows us do define differ-
ential calculus on manifolds.
Note that if f : M N is smooth, for any curve through x M , f is a curve through f(x) N . So we can defineDefinition 2.2.6. The differential of f at x M is
Dxf : TxM Tf(x)N , Dxf([]) = [f ]
The differential is sometimes denoted by f, or dxf when f is a real-valuedfunction.
As this is a generalization of the existing definition of the differential in
Rn, we should check that it actually agrees with it: In analysis, the differentialis usually defined by
Dxf(h) =d
dt
t=0
f(x+ th) , h Rn
In our terms the vector h is the velocity vector of the curve (t) = x + th
and by definition Dxf(h) is the velocity vector of the curve f(x+ th), which
is simply f .Remark. In terms of derivations, Dxf : D 7 Df, where
Df() =d
dt
t=0
(f ) = ddt
t=0
(f ) = D(f )
So, Dxf(D) = D f . In this form it is easier to see that Dxf is a linearmap, as it should be.
As we did with the tangent space, we globalize the concept of the differ-
ential.
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Definition 2.2.7. The differential of f is
Df : TM TN , Df(x, ) = (f(x), Dxf())With the smooth structure on the tangent bundle, we can define smooth
vector fields.
Definition 2.2.8. A smooth vector field on a manifold M is a smooth map
: M TM , x 7 (x, x) , x TxM .By viewing tangent vectors as derivations, we can view a vector field as
an operator on smooth functions:
: C(M) C(M) , (())(x) = x()Thus the smoothness of can be defined in more abstract terms:
Proposition 2.2.9. A vector field is smooth iff for any smooth function
C(M), () is smooth.
2.3 Submersions and immersions
Definition 2.3.1. Let f : M N be a smooth map. f is called an immersion is for all x M , Dxf : TxM Tf(x)N isinjective.
f is called an embedding if it is an immersion and a homeomorphismonto its image.
f is called a submersion if for all x M , Dxf is surjective.The following result is a generalization of theorem 1.2.10. First, we need
another definition.
Definition 2.3.2. Let f : M N smooth. y N is called a regular valueof f if for every x f1(y), Dxf is surjective. Otherwise y is a critical valueof f (note that any y / f(M) is trivially a regular value of f).Theorem 2.3.3 (Preimage Theorem). Let f : Mm Nn be a smooth map,m n, and y f(M) a regular value of f . Then L := f1(y) is a subman-ifold of M of dimension m n. Furthermore, for x L, TxL = kerDxf TxM .
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Proof. As the submanifold condition is local, it is enough to show that L
is a submanifold around all x L. By definition, L = {f y = 0}, andrank(Dxf) = n for all x L. By choosing local coordinates around x, theclaim reduces to theorem 1.2.10 and example 2.1.8.
In particular, if f is a submersion then for any y f(M), f1(y) is asubmanifold. The following theorem is in a sense dual, and justifies the name
embedding.
Theorem 2.3.4. Let f : Mm Nn be an embedding. Then f(M) is asubmanifold of N of dimension m.
Proof. Consider first the case where U Rm is open and f : U Rn is anembedding. Assume that 0 M and f(0) = 0. Well show that f(U) is asubmanifold of Rn near 0.Let k = nm. Denote elements in Rn by (x, y) , x Rm , y Rk, and define
F : U Rk Rn , F (x, y) = f(x) + (0, y)
Since D0f is injective, we may assume its first m rows are linearly indepen-
dent, so that
D0f =
(A
H
)where A Mm(R) is invertible and H is some m k matrix. Then
D0F =
(A 0
H I
)So D0F is invertible, and by the Inverse Function Theorem F is a diffeomor-
phism around 0, and by shrinking U we may assume that F is a diffeomor-
phism on U . Note that by construction, f(U) = F (U {0}). By definition,this implies that f(U) is a submanifold of dimension m around 0.
Now turn to the general case, where f : Mm Nn is an embedding. Letb = f(a) f(M). Choose coordinates around a , b, defined on U M , V N respectively, and note that since f is an embedding, we may assume that
f(U) = V f(M). In local coordinates, the claim reduces to the previouscase.
The question we are trying to answer now is - can any manifold be em-
bedded in Rn?The following example shows that generally the answer is no.
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Example 2.3.5. Define S = S1 {1, 2}. We can picture S as two disjointcopies of the 1-sphere. Denote the north pole by P and define an equivalence
relation
(x, 0) (y, 1) x = y 6= PThus we identify corresponding points in both spheres apart from the poles.
Set M = S/ . M can be pictured as a sphere with two noth poles, P1 andP2. Note that any two neighborhoods of P1 and P2 have non-zero intersection,
so M is not Hausdorff. Since this is a hereditary property, i.e. any subspace
of a Hausdorff space is itself a Hausdorff space, we see that M cannot be
embedded in Rn.
We see that we need to enforce certain restrictions on a manifold for it
to be embedded in Rn. First, we need some results from topology:
Excercise 2.3.6.
(i) Let X be a Hausdorff space and Y X compact. Then Y is closed.(ii) Let X be a compact space and Y X closed. Then Y is compact.Corollary 2.3.7. Let X be compact, Y Hausdorff and f : X Y a contin-uous injection. Then f is a homeomorphism onto its image.
Proof. Since f is already continuous and injective, all that is left is to show
that f1 is continuous. Let F X be closed. By the previous exerciseF is compact. Since f is continuous, f(F ) is also compact. Again, by the
previous exercise f(F ) is closed, proving that f1 is continuous.
Remark. Note that what we have actually proved is that any continuous
map f : X Y with X compact and Y Hausdorff is closed, i.e. f(F ) isclosed for any closed F X.
We can now prove:
Theorem 2.3.8. Any compact Hausdorff manifold can be embedded in RNfor some N .
We start with a
Lemma 2.3.9. Let Bn(r) = {x Rn : |x| < r}. There exists a smoothf : Bn(3) Sn such that
f(Bn(2)
)= Sn \ {P} , f
(Bn(3) \ (Bn(2)) = {P}
and fBn(2)
is a diffeomorphism.
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Proof of the theorem. Let Mn be a compact Hausdorff manifold, and let
{(U, )} be an atlas such that (U) Bn(3) and M =
1(Bn(2)
).
From compactness, we can find an open subcover
M =di=1
Vi , Vi = 1i (B
n(2))
Define
gi : M Rn+1 , gi(x) ={f(i(x)
)x Vi
P otherwise
with f as in the previous lemma. Then giUi
= f i is a diffeomorphism,and gi(x) = P x M \ Vi.
Well show that gi is continuous. Denote Vi = 1i
(Bn(2)
). Note that
giVi
is still continuous, and since i : Ui Rn is a homeomorphism, Vi iscompact.
Now, let F Rn+1 be closed. Consider two cases:(i) If P / F , then g1i (F ) = (gi
V)1(F ) =: Z is by continuity of gi
Va
(relatively) closed subset of Vi, so it is compact. Since M is Hausdorff,
this implies that Z M is closed.(ii) If P F , then g1i (F ) = (g1i (F ) Vi
Z
) (M \ Vi). Note that from
(i), Z = (giV)1(F ) is closed, and M \ Vi is by definition closed. Thus
g1i (F ) is closed.
We see that g1i (F ) is closed, and so gi is continuous.Finally, set N = d(n+1). Define g = (g1, . . . , gd) : M RN . Well check
that g is an embedding:
g is an immersion: We have Dxg = (Dxg1, . . . , Dxgd). On every Vi, gi is
a diffeomorphism, so Dxgi is an isomorphism there, and in particular
injective. Then Dxg is injective there. SinceM =i
Vi, Dxg is injective
throughout M , and g is an immersion.
g is a homeomorphism: On every Vi, gi is injective, so by the same argu-
ment as above, g is injective. We have seen that each gi is continuous,
so g is also continuous, and it is a map from a compact space to a
Hausdorff space. Then g is a homeomorphism onto its image.
So, g : M RN is an embedding.
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3 Boundary and orientation
3.1 Manifolds with boundary
We start by expanding the class of smooth manifolds, allowing them to have
boundary. For instance, the closed ball Dn = {x Rn : |x| 1} is not amanifold by our former definition.
Definition 3.1.1. Let M be a manifold and g : M R be a smooth map,with 0 as a regular value. Then M = {g 0} is called a manifold withboundary, and M = {g = 0} is its boundary.
We give another, more abstract, definition.
Definition 3.1.2. A topological space M is called a (topological) manifold
with boundary if every x M has a neighbourhood homeomorphic to an opensubset of the half-space Hn = {x Rn : x1 0}. A point x M is called aboundary point if its image under such a chart is in Hn = {x Rn : x1 = 0}.Otherwise it is an interior point. The set of boundary points is called the
boundary of M , denoted M .
For a manifold with boundary, we extend the definitions of smooth atlases
and smooth structure.
This definition is a generalization of the definition of a manifold, which
becomes a manifold with boundary with an empty boundary.
Claim. If M is an n-dimensional manifold with boundary with M 6= , Mis a manifold without boundary of dimension n 1.
We turn now to generalize theorem 2.3.3 to manifolds with boundary.
Proposition 3.1.3 (Preimage Theorem for Manifolds With Boundary). Let
Mm be a manifold with boundary, Nn a manifold without boundary and
f : M N a smooth map. Let y N be a regular value for both f andfM
. Then L = f1(y) is a submanifold of M with boundary of dimensionm n, with L = L M , and TxL = kerDxf TxM .
3.2 Orientation
We begin by defining orientation in linear space.
Let V = Rn be an n-dimensional real linear space. We denote by L(V )the (linear) space of all linear maps V V . Let B1 , B2 be two bases, and
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let A L(V ) be the transition matrix between them. Then det(A) 6= 0.We say that B1 and B2 have the same orientation, denoted B1 B2, ifdet(A) > 0. It is clear that this equivalence relation admits exactly two
equivalence classes, called orientations of V .
To extend this definition to manifolds, we first find an equivalent and
more abstract definition for orientation on linear spaces.
Definition 3.2.1. A linear skew-symmetric n-form (in short: n-form) on V
is map : V . . . V n times
R, linear in all arguments and skew-symmetric
under changing any two variables, that is
(v1, . . . , vi, . . . , vj, . . . , vn) = (v1, . . . , vj, . . . , vi, . . . , vn)
The set of all n-forms on V is denoted n(V ).
Note that n(V ) is a linear space over R. Moreover, we have:
Excercise 3.2.2. For any n(V ) and A L(V ),(Av1, . . . , Avn) = det(A) (v1, . . . , vn)
As an immediate consequence we get the following useful
Corollary 3.2.3. dim(n(V )
)= 1
Proof. Let e1, . . . , en be some basis for V . For any v1, . . . , vn V , if the viare linearly dependent, (v1, . . . , vn) = 0 for all n(V ). Otherwise wecan find a unique nonsingular P L(V ) such that
Pei = vi 1 i nThen for any n(V ),
(v1, . . . , vn) = det(P ) (e1, . . . , en)We see that all n-forms differ by a scalar factor.
We now arrive at the fundamental definition of this section.
Definition 3.2.4. n(V ) is called a volume form if it is non-zero.If 1 , 2 are volume forms,
12
is well-defined and non-zero. We say that
1 and 2 are equivalent if12
> 0.
18
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Definition 3.2.5. An orientation on V is an equivalence class of volume
forms.
If is a volume form and B = {e1, . . . , en} an ordered basis, we say that respects B, or B is positively oriented with respect to if (e1, . . . , en) > 0.
Now turn to manifolds. Since orientation is defined on linear spaces, it is
natural to define orientation on manifolds in term of the tangent space.
Definition 3.2.6. Let Mn be a manifold. An n-form on M is a smooth
assignment x 7 x n(TxM). The set of smooth n-forms onM is denotedn(M).
The smoothness of can be defined as follows. Choose coordinates
around a M , then TxM Rn for x near a. Then there is a naturalvolume form on TxM , the determinant, so any n-form can be represented
around a as x = f(x) det. We say that is smooth around a if f issmooth. We say that is smooth if it is smooth around all a M .
As with vector fields, the smoothness of volume forms can be defined
equivalently and more abstractly - we say that is smooth if for any smooth
vector fields 1, . . . , n, (1, . . . , n) is a smooth function on M .
Definition 3.2.7. A volume form on M is a non-vanishing n(M).If 1 , 2 are two volume forms on M , then
12
is a smooth non-vanishing
function on M . We say that 1 and 2 are equivalent if12
> 0.
Definition 3.2.8. An orientation on M is an equivalence class of volume
forms. M is called orientable if there exists a volume form on M . (M, [])
is called an oriented manifold.
Excercise 3.2.9. Let M be an orientable manifold with k connected com-
ponents. Then M admits 2k different orientations.
We now extend the notion of pullback to volume forms.
Definition 3.2.10. LetMn and Nn be manifolds, and f : M N a smoothmap. Define the pullback map:
f : n(N) n(M) , f x(1, . . . , n) = f(x)(Dxf(1
), . . . , Dxf(n))
Remark. Note that for f to pull back volume forms to volume forms, Dxf
must be an isomorphism, i.e. f must be a local diffeomorphism.
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Claim.
(fg) = gf
For C(N) , n(N) , f () = f f Example 3.2.11. The torus Tn.
Claim. Tn is orientable.
Proof. Denote 0 = det the natural volume form on Rn, pi : Rn Tnthe natural projection. Let x Rn , y Tn such that y = pi(x). Thenpi : TxRn TyTn is an isomorphism. Define
(1, . . . , n) = 0(pi1 1, . . . , pi
1 n)
This form is well-defined, since 0 is translation-invariant.
Example 3.2.12. Sn Rn+1. Let nx be the unit normal to Sn at x, and 0as before. Define 0 = inx0 , where:
inx0(1, . . . , n) = 0(nx, 1, . . . , n)
Then 0 is a volume form on Sn. As we will see later, this example can be
generalized. The operator inx is called interior product with the vector nx
Example 3.2.13. The real projective space RP n.
Claim. RP n is orientable iff n is odd.
Proof. Suppose RP n is orientable, and let be a volume form. Let pi : Sn RP n be the natural projection, and 0 the standard volume form on Sn asdefined earlier. Define f : Sn Sn , f(x) = x. Note that by definitionpi f = pi, so pi = f pi. Define = pi, a volume form on Sn. Then
f = f pi = pi =
, 0 are two volume forms on Sn, so we have
= 0 ( 6= 0) = 0 = f = f (0) = f f 0 f
00
=
f > 0 (i)
20
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Since f (x) = (x), and as is non-vanishing it has constant sign on Sn.Now, f = , and nf(x) = nx = nx. So,
f 0(v1, . . . , vn) = 0(v1, . . . ,vn) =0(nx,v1, . . . ,vn) = (1)n+10(v1, . . . , vn)
f00
= (1)n+1 (ii)
From (i) and (ii) we get an immediate contradiction if n is even. If n is
odd, however, we see that f 0 = 0, which allows us to define a volumeform on RP n - if y = pi(x), x Sn, set
y(v1, . . . , vn) = 0x(pi1 v1, . . . , pi
1 vn)
By our observation, is well-defined.
Example 3.2.14. The Mobius strip: Let
f : R2 R2 , f(x, y) = (x+ 1,y)Define the orbit of a R2, Oa := {fk(a) : k Z}. Define an equivalencerelation by a b b Oa, and set M = R2/ . M is called the Mobiusstrip. It can be shown, very similarly to 3.2.13, that M is not orientable.
Definition 3.2.15. Let (M,M) and (N,N) be two orientable manifolds of
the same dimension and f : M N a smooth map. We say that f preservesorientation with respect to M , N if f
N M .Remark. This is essentially a local condition, as we have seen in 3.2 that
f preserves the non-vanishing of a form if f is an isomorphism, which is a
local condition. Clearly, iff NM
> 0 in x, it holds in a neighbourhood of x.
Example 3.2.16. If F : Rn Rn is smooth and Rn is considered with thestandard orientation, then F preserves orientation around x iff JF (x) :=
detDxF > 0.
The presence of the standard volume form on Rn allows us to give anequivalent definition of orientable manifold, similar to the construction of a
smooth structure.
Definition 3.2.17. LetM be a manifold. An atlas onM is called orientable
if its transition functions all preserve the standard orientation in Rn.M is called orientable if it admits an orientable atlas.
To see the equivalence between these definitions, we need the partition of
unity, discussed in the next section.
21
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3.3 The Partition of Unity
We start with a few definitions from topology. Let X be a topological space.
Definition 3.3.1. A collection {U} of subsets of X is called a cover if
U = X.
Definition 3.3.2. A cover {U} of X is called locally finite if for any x Xthere is a neighbourhood x V such that V U = for all but finitelymany U.
Definition 3.3.3. A cover {U} is said to be a refinement of a second cover{V} if for every there is a such that U V.Definition 3.3.4. A topological space is called paracompact if every open
cover admits a locally finite open refinement.
We borrow the following result from topology:
Theorem 3.3.5. Every second countable Hausdorff topological manifold is
paracompact.
From now on, we assume that all manifolds are second countable and
Hausdorff, and therefore paracompact.
Definition 3.3.6. Let f C(M). Define the support of f :supp f = cl{x M : f(x) 6= 0}
We denote the subset of smooth functions with compact support Cc (M).
Theorem 3.3.7 (Partition of Unity). LetM be a manifold and {U} an opencover of M . Then there exists a family of smooth functions {f : M [0, 1]}which satisfies
1. supp f U.2. The collection {supp f} is locally finite.3.
f = 1 throughout M .
Note that condition 2 ensures that the sum in 3 is everywhere finite, so
that condition 3 is meaningful.
The family {f} is called a partition of unity subordinate to the opencover {U}.
We use the following construction, for which we do not give a complete
proof.
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Proposition 3.3.8 (The cutoff function). Let U be a coordinate neighbour-
hood in M , and V U an open set such that V U and is compact. Thenthere exists a smooth function f : M [0, 1] with compact support such that
1. f(x) = 1 for all x V .2. f(x) = 0 for all x / U .
Such a function is called a cutoff function for V U .Proof. The main steps towards a proof are:
First find a function f : R R which satisfies f(t) = 1 for t a andf(t) = 0 for f b. The idea is to find a smooth function g vanishingoutside [a, b] and then use the function f(t) =
tg(s)ds, normalizing
if necessary.
Find a cutoff function for concentric ball in Rn, by setting F (x) =f(|x|).
Find a cutoff function for open sets V V U Rn with V compactby covering V by finitely many open balls.
Finally, define the cutoff function on M using local coordinates.
Proof of the Partition of Unity theorem. Again, we give only a sketch of a
proof:
By paracompactness and second countability we may assume that {U} iscountable and locally finite, and that each U is a coordinate neighbourhood.
Next, we can find an open cover {V} such that V U.Find cutoff functions for V U, then :=
is a locally finite
sum and positive throughout M .
Finally, set f =.
We can now prove that the two definitions given for orientability are
equivalent.
Theorem 3.3.9. Let M be a boundaryless manifold. The following condi-
tions are equivalent:
(i) M admits an orientable atlas.
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(ii) M admits a volume form.
Proof. (i) (ii) Denote by 0 the standard volume form on Rn. Let{(U, )} be an orientable atlas on M . Define = 0, then is avolume form on U. By definition, whenever U U 6= , T = 1preserves orientation. Then = T preserves orientation with respectto and 0 , so
> 0 where it is defined. Now, let {h} be a partitionof unity subordinate to the open cover {U}. Define =
h. Then
is an n-form on M , and near every x M we have = ki=1 hii , afinite sum, and we can assume WLOG that hk(x) > 0. Since supphk Uk,x Uk, where k is non-vanishing. Then near x, we have
=ki=1
hii =
( ki=1
hiik
>0
)k
Thus is a volume form on M .
(ii) (i) Suppose is a volume form on M, and let {(U, )} be someatlas on M. For each , if is not orientation preserving, we can modify it
to become so, for instance by setting = (x1, . . . , xn) = (x1, x2, . . . , xn).This clearly does not change the smoothness of the transition functions, and
so we can construct a new atlas {(U, )} whose coordinate functions areall orientation preserving. Such an atlas is clearly orientable.
In order to address orientation on manifolds with boundary, we first dis-
cuss metrics on manifolds.
3.4 Riemannian manifolds
Definition 3.4.1. Let M be a manifold. A Riemannian metric on M is a
symmetric, positive-definite bilinear form (or inner product) on TM , i.e a
smooth assignment B : x 7 Bx, where Bx is such a form on TxM . (M,B) iscalled a Riemannian manifold.
Again, the smoothness of B can be defined in two ways. By choosing local
coordinates around a we have TxM Rn near a, so we have the Euclideanform on TxM . As is known from linear algebra, any inner product can be
expressed as
B(u, v) = Beuc(u,Av)
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where A is a positive-definite linear map. So B defines a map x 7 Ax, andwe say that B is smooth if this map is smooth.
More abstractly, we say that B is smooth if for any two smooth vector
fields , , B(, ) is a smooth function.
Theorem 3.4.2. A Riemannian metric exists on any manifold.
Proof. Let {(U, )} be a locally finite atlas on M , and h a compatiblepartition of unity. On each U, the local coordinates give an isomorphism
TxM Rn. Define B as the Euclidean form in local coordinates, i.e.B(u, v) = Beuc(u, v)
Then B is a metric on U. Define B =
hB. Then B is a symmetric
bilinear form on TM . Let x M , then near x, B = ni=1 hiBi, withx supphn Un. As we have seen, any other form can be expressed asBi(, ) = Bn(, Ai) with Ai a positive-definite linear operator. Then aroundx,
B(, ) =ni=1
hiBi(, ) =ni=1
hiBn(, Ai) = Bn
(,
ni=1
hiAi
)Clearly,
ni=1 hiAi is a positive-definite linear operator. Thus B is a Rie-
mannian metric on M .
Proposition 3.4.3. Let (M,B,) be an oriented Riemannian manifold with
boundary. Then M is orientable.
Proof. We use the first definition of a manifold with boundary, i.e. M issome manifold, f : M R is smooth and M = {f 0}. Let x M . ThenTxM TxM is of codimension 1. Let nx TxM such that
1. nx TxM with respect to B.2. ||nx|| :=
B(nx, nx) = 1.
3. dxf(nx) > 0.
Clearly, such a vector exists, and the map x 7 nx is smooth. nx is called theoutward unit normal toM at x. Define a form on M by = inx. Since nxis non-vanishing, is a volume form on M , called the induced orientation
on M .
Combining the last two results, we conclude:
Theorem 3.4.4. Let M be an orientable manifold with boundary. Then M
is orientable.
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4 Homotopy and degree of a map
4.1 Homotopy
Definition 4.1.1. Let f , g : M N be smooth maps. We say that f andg are homotopic, denoted f g, if there exists a smooth : M [0, 1] Nsuch that (x, 0) f(x) , (x, 1) g(x).Example 4.1.2. Every f : M Rn is homotopic to the trivial map g 0by (x, t) = t f(x).
An important case of homotopy is on the n-sphere.
Theorem 4.1.3. Let : Sn Sn be the antipodal map, (x) = x. Then is homotopic to the identity map iff n is odd.
One direction follows from the next theorem. We will prove the second
direction later, when we have developed the proper tools.
Theorem 4.1.4. If there exists a non-vanishing vector field v on Sn then
the antipodal map is homotopic to the identity map.
Proof. Since v is non-vanishing we can assume it is normalized, i.e. ||v(x)|| =1 for all x Sn. Thus we can view it as a map from the sphere onto itself,x 7 v(x). Note that the normal to Sn at x is x itself, so v(x) x. Define
: Sn [0, pi] Sn , (x, ) = x cos + v(x) sin
Computation yields ||(x, )|| = 1 for all (x, ) and so is indeed into Sn.In addition, (x, 0) = x , (x, pi) = x = (x), as required.Corollary 4.1.5. For odd n, the antipodal map is homotopic to the identity.
Proof. Let S2k1 R2k. As noted above, TxSn = {nx} = {x}. Define
: Sn TxSn , (x1, . . . , x2k) 7 (x2, x1, . . . ,x2k, x2k1)
Then is a smooth non-vanishing vector field on Sn, so is homotopic to
the identity.
Theorems 4.1.3 and 4.1.4 together imply
Theorem 4.1.6 (Porcupine Theorem). There is no non-vanishing vector
field on S2n.
26
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Theorem 4.1.7. The identity and constant maps on Sn are not homotopic.
Again, we will prove this theorem later. In the meantime, we can find
some important corollaries.
For the following, recall the notation Dn = {x Rn : |x| 1}.Corollary 4.1.8. There is no smooth retract from Dn+1 to Sn, i.e. a smooth
F : Dn+1 Sn such that F Sn
= 1 - the identity map.
Proof. Suppose that F is such a retract. Define
: Sn [0, 1] Sn , (x, t) = F (tx)
Then (x, 1) = F (x) = x, and (x, 0) F (0), contradicting the previoustheorem.
Corollary 4.1.9 (Brouwer Fixed-Point Theorem). Every smooth map f : Dn Dn has a fixed point, i.e. there exists x Dn such that f(x) = x.Proof. Suppose otherwise, i.e. f(x) 6= x for all x Dn. Then we canconstruct a smooth retract F : Dn Sn1 as follows:
Figure 2: Brouwer Fixed-Point Theorem
This contradicts the previous corollary, so f must have a fixed point.
Remark. The Brouwer Fixed Point theorem is actually more general, stated
for continuous functions. It can be reduced to the smooth case by using the
Stone-Weierstrass theorem which states that the smooth functions are dense
in the continuous functions with respect to the sup norm, ||f || = supxDn
|f(x)|.
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4.2 The Degree Theorem (I)
We keep the following assumptions throughout this section: Mn, Nn are
oriented manifolds of equal dimensions, M is compact and boundaryless and
N is connected.
Proposition 4.2.1. Let f : Mn Nn be a smooth map, and y N be aregular value of f . Then f1(y) is finite.
Proof. Assume f1(y) 6= . Since f is continuous, f1(y) is closed in M , soit is compact. We have seen (in theorem 2.3.3) that it is a submanifold of
dimension nn = 0. Such a manifold is a set of isolated points, as each suchpoint has a neighbourhood homeomorphic to a point. These neighbourhoods
form an open cover of f1(y), so by compactness it follows that there arefinitely many such points.
Definition 4.2.2. Let y be a regular value of f , then as we have just seen
f1(y) is finite. For x f1(y) define
(x) =
{1 f preserves orientation around x
1 otherwiseAnd define the degree of f over y as
d(f, y) =
xf1(y)(x)
The main result in this section is the following important theorem:
Theorem 4.2.3 (Degree Theorem).
1. d(f, y) does not depend on the choice of a regular value y. Then we can
define deg(f) = d(f, y) for some (and all) regular value.
2. If f , g : M N are homotopic then deg(f) = deg(g).Remarks.
1. If f : M M , deg(f) does not depend on the choice of orientation onM - indeed, if f : M N , then deg(f) does not change under reversingorientation on both manifolds.
2. In particular, if f : M M is a diffeomorphism, any y M is a regularvalue and f1(y) contains a single point, so deg(f) = 1.
28
-
Before proving the theorem, we demonstrate its power with some imme-
diate results.
Example 4.2.4. We have seen that the antipodal map : Sn Sn pre-serves orientation iff n is odd, therefore deg() = (1)n+1. Since obviouslydeg(1) = 1, we conclude that the identity and antipodal maps are not ho-
motopic for even n, the second half of theorem 4.1.3.
Example 4.2.5. The constant map f : M M , f(x) x0. Then x M isa regular value iff x 6= x0, and then f1(x) = . Therefore deg(f) = 0, andso the constant map is not homotopic to the identity map, proving theorem
4.1.7.
The following two results are stated without proof. Their proofs can be
found in Milnors book [2].
Theorem 4.2.6 (Morse-Sard Theorem). Let f : M N be a smooth map.Then the set of regular values of f is dense in N .
Theorem 4.2.7 (Classification of One-Manifolds). Any compact, connected
1-dimensional manifold is diffeomorphic to S1 or [0, 1].
We give the proof of the Degree Theorem in the form of several lemmas:
Lemma 1. Suppose Mn = Xn+1, with X compact, F : X N a smoothmap and f = F
M. Let y N be a regular value of both F and f . Then
d(f, y) = 0.
Proof. Denote = F1(y). Since F is continuous, is closed, so it is acompact 1-dimensional submanifold of X, and = f1(y). By theorem4.2.7 any connected component of is either boundaryless or has exactly
two boundary points, denote them x. We will show that for any such pair(x+) + (x) = 0. Therefore we can assume WLOG that [0, 1], andf1(y) contains exactly two points. Choose a parametrization
: [0, 1]M , (0) = x , (1) = x+ , 6= 0
For all t, F((t)
)= 0. Since f = F
M, f = F
TxM
for x M . As y is aregular value of f , we have
(i) rank(FTxM
) = rank(f) = n = dimTyN
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Thus FTxM
is injective. Therefore, (t) / T(t)M for all t [0, 1]. Inparticular, / TxM . This allows us to choose a Riemannian metric on Xsuch that (0) TxM , (1) Tx+M . Define
Et = {(t)} T(t)MBy (i), F : Et TyN is an isomorphism of linear spaces. Choose a basis(e1, . . . , en) for TyN and set
vi = (FEt)1(ei)
Then (v1, . . . , vn) is a basis for Et. Let be a volume form on X, and M =
inx the induced orientation on M . Then (, v1, . . . , vn) has constant sign
throughout . Note that up to a positive scalar factor, nx = (0) , nx+ =(1). Therefore
M(v1(0), . . . , vn(0)
) (nx , v1(0), . . . , vn(0)) (nx+ , v1(1), . . . , vn(1)) M(v1(1), . . . , vn(1))
where means of the same sign. Obviously,f N
(v1(0), . . . , vn(0)
)= f N
(v1(1), . . . , vn(1)
)So, (x) + (x+) = 0, hence d(f, y) = 0.
Lemma 2. The same result holds without the assumption that y is a regular
value of F .
Proof. We will prove a slightly stronger result, since we will use it later: We
will show that the set of regular values of f is open, and that d(f, y) is a
locally constant function of y. That is, for any regular value y N we canfind a neighbourhood U such that any z U is a regular value of f withd(f, z) = d(f, y). Combining this with theorem 4.2.6, the lemma follows
easily.
Indeed, since dimM = dimN and y is a regular value, for any x f1(y)there exist open sets x V , y U such that f
V: V U is a diffeomor-
phism. Moreover, by shrinking V we may assume thatf NM
has constant
sign on V . As f1(y) is finite there exist finitely many such neighbourhoodsVi, 1 i n, which cover f1(y). Set
U =ni=1
Ui
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Then U is almost the required neighbourhood, but for z near y there we
might still have points in f1(z) outside all Vi, of which we have no informa-tion. To fix this, set
V =ni=1
Vi
Since M is compact and N is Hausdorff, f is a closed map (see corollary
2.3.7). ThenW := f(M\V ) is closed. Finally, set U = U\W , an open subsetof N with f1(U) V . Then each point in U has exactly one preimage ineach Vi, all with similar orientation, so U is the required neighbourhood.
Lemma 3. Let f , g : M N be homotopic smooth maps and y N aregular value of both f and g. Then d(f, y) = d(g, y).
Proof. Let be a volume form on M . Denote X := M [0, 1], then X hasa natural product orientation. Let F : X N be a homotopy between f , g.As an oriented manifold,
X = (M {0},) unionsq (M {1}, )Now,
FX
(x, t) =
{f(x) , t = 0
g(x) , t = 1
so y is a regular value of FX
. By the previous lemma
0 = d(FX, y) =
xf1(y)
F (x) +
xg1(y)F (x)
=
xf1(y)(1) f (x) +
xg1(y)
g(x) = d(g, y) d(f, y)
Or, d(f, y) = d(g, y).
Lemma 4. Let y, z N be regular values of f . Then d(f, y) = d(f, z).The last lemma is in fact the first part of the Degree Theorem. We
postpone its proof until later, and first use it to proof the second part of the
theorem:
Proof of the Degree Theorem. Let f, g : M N be homotopic maps. Bylemma 4, deg(f) and deg(g) are well-defined. Denote by Xf , Xg their respec-
tive sets of regular values. By lemma 2 and theorem 4.2.6 these are dense
open subsets of N . Thus Xf Xg 6= . Take w Xf Xg. Thendeg(f) = d(f, w) = d(g, w) = deg(g)
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This proves the second part and completes the proof of the Degree Theorem,
not including lemma 4.
We now turn to prove the crucial lemma. To do so we need to introduce
a new concept:
Definition 4.2.8. Let N be a manifold. A smooth isotopy is a smooth
h : N [0, 1] N such that h0 = 1 and for all t, ht : N N is a diffeomor-phism.
Define the support of h, supph = cl{x N : t. ht(x) 6= x}.Definition 4.2.9. x, y N are called isotopic if there is an isotopy h on Nsuch that h1(x) = y.
Excercise 4.2.10. This is an equivalence relation.
Lemma 4.2.11. Any x, y Bn = {x Rn : |x| < 1} are isotopic throughan isotopy with compact support.
Proof. Consider the constant vector field 0(w) = y x defined on the com-pact interval
[x, y] := {x+ (1 )y : [0, 1]}Using a cutoff function we can extend it to a smooth vector field defined
on all of Bn satisfying (w) = y x in an open neighbourhood of [x, y] and(w) = 0 outside a compact K Bn.
Consider the ODE
w(t) = (w) , w(t) Bn, t R+
From the theory of differential equations we know that for any initial condi-
tion w0 there is a unique solution w(t). Moreover, the map ht : w0 7 w(t)defines an isotopy on Bn. Clearly, taking the initial condition w0 = x we
get w(1) = y. Since (w) = 0 for w / K, ht(w) = w for all t for such w,and thus supph K. So the isotopy h has compact support, and satisfiesh1(x) = y.
Corollary 4.2.12. Let N be a connected manifold. Then any x, y N areisotopic.
Proof. First note that for any x N there is an open neighbourhood V suchthat any y V is isotopic to x. For, given local coordinates near x, using the
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previous lemma we can construct an isotopy with compact support satisfying
h1(x) = y, and then extend it to all of N using a cutoff function.
We have seen that isotopy between points is an equivalence relation. By
our previous observation, each equivalence class is open, and so we have a
partition of N into open disjoint sets. Since N is connected, there can be
only one such equivalence class.
We can now restate and prove:
Lemma 4. Let y, z N be regular values of f . Then d(f, y) = d(f, z).Proof. Let h be an isotopy on N such that h1(y) = z. For all t, ht : N Nis a diffeomorphism, so ht(y) is a regular value of ht f .
From similar reasoning to that of lemma 2 it follows that d(ht f, ht(y))is a locally constant function of t, and as it is defined on the connected set
[0, 1] it is constant. Setting t = 0 and t = 1 we get d(f, y) = d(h1 f, z).Since h1 f is clearly homotopic to f and z is a regular value of both,
from lemma 3 we get d(h1 f, z) = d(f, z). Combining the two, we getd(f, y) = d(f, z).
4.3 Euler characteristic and vector fields
Let U Rn be an open subset and v a vector field on U . Let x0 be a zeroof v, i.e. v(x0) = 0. x0 is called a non-degenerate zero if det
(ivj(x0)
) 6= 0.More accurately, on Rn we have the natural isomorphism TxRn Rn. Thenwe identify the vector field v(x) = (x, v(x)) with the map v : Rn Rn, andthen x0 is a non-degenerate zero if Dx0v is invertible.
Lemma 4.3.1. This definition does not depend on choice of coordinates.
Proof. Assume WLOG x0 = 0. Let F : U U , F (0) = 0, be a diffeomor-phism corresponding to the change of coordinates.
In the new coordinates, v is given by:
v = (F1)v = DxF v F1This section is not complete - most proofs are sketched and some are omitted alto-
gether. For further details see Milnors book [2]
33
-
Denote A(x) := DxF . Then from the smoothness of F it follows that A(x) =
A(0) + o(||x||). Let h Rn. Differentiating both sides at x = 0, we get:
DF (0)v D0F h = ddt
t=0
A(0 + th) v(0 + th) =
=d
dt
t=0
(A(0) + o(1)
) (v(0)=0
+tD0v h+ o(t))=
=d
dt
t=0
(tA(0) D0v h+ o(t)
)=
= A(0) D0v h = D0F D0v h
So, DF (0)v = D0F D0v (D0F )1. Since D0F is an isomorphism, the non-degeneracy of the zero does not depend on coordinates.
This allows us to define:
Definition 4.3.2. The index of v at x0 is:
indx0(v) = sgn det(Dx0v)
Claim. Every non-degenerate zero is isolated.
Proof. Let x0 be such a zero. Then Dx0v is invertible, so the Inverse Func-
tion Theorem implies that v is injective in some neighbourhood of x0. In
particular, there are no other zeros there.
This suggests examining general isolated zeros. Let x0 be such a zero,
not necessarily non-degenerate. Then we have a ball B around x0 where v
does not vanish. Denote S = B.
Definition 4.3.3. Consider the map:
g : S Sn1 , x 7 v(x)||v(x)||We define the index of v at x0 as:
indx0(v) = deg(g)
This definition does not depend on the choice of coordinates, or of , since
we can always find a homotopy between two such spheres (See Milnors book
[2] for details).
34
-
Remark. Note that this index is invariant under a homotopy which leaves
x0 an isolated zero. Indeed, if vt is such a homotopy between v0 and v1, it
induces a smooth homotopy between g0 and g1, and thus deg(g0) = deg(g1).
Example 4.3.4. Let C R2, v : z 7 zn. The only zero of v is at 0. Identify{z C : |z| = 1} S1. Then
g : S1 S1 , 7 n
Thus ind0(v) = deg(g) = n.
Proposition 4.3.5. For a non-degenerate zero, the two definitions coincide.
Proof. First, consider the case where v(x) = Qx, with Q an orthogonal
matrix. Then on S, ||Qx|| = ||x|| = and so,
g : S Sn1 , x 7 1Qx
Then,
deg(g) = sgn detQ = sgn detDv
Next, consider v(x) = Ax where A is some nonsingular matrix. We use the
polar decomposition A = PQ, where P is symmetric and positive-definite
and Q orthogonal. In particular, detP > 0. Define a family of vector fields:
vt(x) = RtQx :=(t I + (1 t)P)Qx
Note that for all t, vt has a non-degenerate zero at 0 and Rt is symmetric
and positive-definite. This defines a smooth homotopy between v = v0 and
v1(x) = Qx, and then
ind0(v0) = ind0(v1) = sgn detQ = sgn detPQ = sgn detA
Finally turn to the general case - Let 0 be a non-degenerate zero of v. Then
v(x) = D0v x+R(x) , R(x) = o(||x||)
Again, define the family:
vt(x) = D0v x+ (1 t)R(x)
This defines a homotopy between v = v0 and v1(x) = D0v x and reducesthe claim to the previous case.
35
-
We give the following theorem without proof:
Theorem 4.3.6. Let M be a closed (=compact and boundaryless) manifold.
Then:
1. There exists a vector field on M without degenerate zeroes.
2. For such a field, define:
ind(v) =
v(x)=0
indx(v)
Then ind(v) is the same for any such vector field v.
Definition 4.3.7. Let M be a closed manifold. Define its Euler character-
istic: (M) = ind(v).
Let Mm Rn be a closed manifold. Define its tubular neighbourhood:
N = {x Rn : dist(x,M) }
and set = N
Excercise 4.3.8. For sufficiently small , N is an n-dimensional manifold
with boundary, n1.
Definition 4.3.9. Denote by (x) the outward unit normal at x . Definethe Gauss map:
G : Sn1 , x 7 (x)Then we have the following important result, which we will not prove:
Theorem 4.3.10. (M) = deg(G)
Excercise 4.3.11. Let f : Mn Sn,M closed. Then deg(f) = (1)n+1 deg(f).Corollary 4.3.12. If dimM is odd then (M) = 0.
Proof. Let v be a vector field on M with no degenerate zeros. Consider the
field v. Clearly their zeros coincide. Let x0 be such a zero. Then we havethe maps
gv, gv : Sn1 Sn1
Now, gv = gv, therefore deg(gv) = deg(gv) and indx0(v) = indx0 v.This holds for all zeros, so (M) = ind(v) = ind(v) = (M).
36
-
Corollary 4.3.13. Let M2k R2k+1. Consider the Gauss map
G : M S2k , x 7 (x)
Then (M) = 2 deg(G).
Proof. Let N, as before. Since codimM = 1, that is,M is a hypersurface,
has two connected components, both of which are diffeomorphic to M .
Moreover, this diffeomorphism is orientation preserving for one component
and orientation reversing for the other. Denote these components, M+,M,respectively. Let
G : M S2k
be the corresponding Gauss maps. By theorem 4.3.10,
(M) = deg(G+) + deg(G)
But by our previous remark,
deg(G+, orM+) = deg(G, orM)
deg(G, orM) = deg(G,orM)
From exercise 4.3.11 it follows that:
deg(G,orM) = deg(G, orM) = deg(G, orM)
Therefore,
(M) = deg(G+) + deg(G) = deg(G) + deg(G) = 2 deg(G)
Example 4.3.14. On the 2-sphere (x) = x, so the Gauss map is G = 1.
Then (S2) = 2 deg(1) = 2.Example 4.3.15. On the torus Tn one can easily construct a non-vanishingvector field - for instance, the field v(x) = 1. Thus (Tn) = 0.
Example 4.3.16. Let M2 R3 be a surface of genus g - that is, a surfacewith g handles. Let G : M S2, the Gauss map, and e = (0, 0, 1) S2.Consider G1(e) (see figure 4). It contains g + 1 points, one of them a localmaxima and the rest saddle points.
37
-
Figure 3: Surface of genus g = 4
Around each p G1(e), M can be represented as a graph {z = f(x, y)}.Then the (unnormalized) normal vector at x is N(x) = (fx,fy, 1). So:
G : (x, y) 7 11 + fx
2 + fy2(fx,fy, 1)
Choose local coordinates x, y on S2 near e, then is the local representation:
G : (x, y) 7 11 + fx
2 + fy2(fx,fy)
Since at all p G1(e), fx = fy = 0, computation yields:
DpG =
(fxx fxyfxy fyy
)
Then, as is known from calculus,
sgn detDpG = fxxfyy fxy2 ={+1 p is a local extremum
1 p is a saddle point
38
-
So, deg(G) = 1 + g (1) = 1 g. Then, from theorem 4.3.10, we get:(M) = 2 deg(G) = 2 2g.
39
-
5 Exterior differential calculus
5.1 Exterior forms on linear spaces
Let V be a real linear space. For k N we denote by Sk the set of permuta-tions on {1, . . . , k}. For Sk we define the sign of ,
sgn =
{1 is an even permutation
1 is an odd permutationWe use the notation =
((1), . . . , (k)
).
Definition 5.1.1. A skew-symmetric linear form, or exterior form, of degree
k, is a map : V . . . V k times
R, linear in all arguments and skew-symmetric
under changing any two variables. We denote by k(V ) the set of all exterior
k-forms, which is a real linear space.
Claim. For any k(V ) and Sk(v1, . . . , vk) = sgn()(v(1), . . . , v(k))
Some simple examples are:
Example 5.1.2. By convention 0(V ) = R.
Example 5.1.3. If k = 1, the skew-symmetry condition is trivial, therefore
1(E) = E.
Example 5.1.4. If dimV = n, then we have seen n(V ) R.Note that from skew-symmetry it follows that for any k(V ) and
v1, . . . , vk V , if vi = vj for i 6= j then (v1, . . . , vk) = 0.More generally, if the vi are linearly dependent, then we get the same result.
It follows that
Claim. Let dimV = n and m > n. Then m(V ) = {0}.Proof. Let any v1, . . . , vm V . Sincem > n they must be linearly dependent,and so for any m(V ), (v1, . . . , vm) = 0.Definition 5.1.5. Define the exterior product (or wedge product):
: k(V ) l(V ) k+l(V ) (v1, . . . , vk+i) =
=(i1,...,ik,j1,...,jl)i1
-
For simplicity, from now on we will simply write summation on assum-
ing we sum only on legal permutations.
Claim. The wedge product satisfies:
1. Linearity: (a + b ) = a + b .2. Associativity: ( ) = ( ).3. Leibnitz Rule: for k(V ) , l(V ), = (1)kl .
Proposition 5.1.6. Let 1, . . . , k 1(V ). Then for any v1, . . . , vk Vwe have:
1 . . . k(v1, . . . , vk) = det(i(vj))Proof. By induction on k. The case k = 1 is trivial. Expand the determinant
by the first row, and use the fact that the appropriate minors are (k 1)(k 1):
det(i(vj)) =kj=1
(1)j+11(vj)Mij = (by induction assumption)
=kj=1
(1)j+11(vj) 2 . . . k(v1, . . . , vj, . . . , vk) = ()
Here vj means that vj is missing. Note that
sgn := sgn(j, 1, . . . , k) = (1)j+1
Then we get:
() =
sgn()1(v(1)) 2 . . . k(v(2), . . . ,(k) ) =
= 1 . . . k(v1, . . . , vk)
Note that from the last result it follows that 1 . . . k = 0 wheneveri = j for i 6= j. For then we have two identical rows in the matrix (i(vj))so its determinant vanishes.
This result is especially useful because we can express all forms in terms
of 1-forms:
41
-
Proposition 5.1.7. Let n = dimV , and let 1, . . . , n be a basis for 1(V ) =
V . Then A = {i1 . . . ik}ii
-
since there must be at least one 1 r k such that (jr) 6= ir.A similar computation yields J(ej1 , . . . , ejk) = 1
So we get:
0 =
(I
I I)(ej1 , . . . , ejk) = J
Thus J = 0 for all J and A is a linearly independent set.
Corollary 5.1.8. Let n = dimV . Then dimk(V ) =(nk
).
Remark. When working with exterior forms, the standard bases in Rn and(Rn) are usually denoted i and dxi, 1 i n, respectfully. We will soongive meaning to these notations, but in the meantime we can consider them
symbolic conventions.
Example 5.1.9. Consider the two-dimensional phase space, i.e. R4 withcoordinates q1, p1, q2, p2. Define the symplectic form:
= dp1 dq1 + dp2 dq2
Then
= (dp1 dq1 + dp2 dq2) (dp1 dq1 + dp2 dq2) == dp1 dq1 dp2 dq2 + dp1 dq1 dp1 dq1
=0
+
+ dp2 dq2 deg=2
dp1 dq1 deg=2
+ dp2 dq2 dp2 dq2 =0
=
= dp1 dq1 dp2 dq2 + (1)4dp1 dq1 dp2 dq2 == 2 dp1 dq1 dp2 dq2 = 2 0
Where 0 is the standard volume form.
Excercise 5.1.10. In the n-dimensional phase space, define the symplectic
form:
= dp1 dq1 + . . .+ dpn dqnThen
. . . n times
= n! 0
Definition 5.1.11. Let V, U be linear spaces, A : V U a linear map.Define its dual map:
A : k(U) k(V ) , A(v1, . . . , vk) = (Av1, . . . , Avk)
43
-
5.2 Differential forms in Rn
Definition 5.2.1. Let U Rn be an open set. an exterior differential formof degree k is a smooth assignment
x 7 x k(TxU)
The set of differential k-forms is denoted k(U).
For k = 0, we have seen that an 0-form on a linear space is a scalar. Thus
a smooth 0-form is simply a smooth function, i.e. 0(U) = C(U).
As usual, the smoothness of a differential form can be defined using local
coordinates, or by its action on smooth vector fields.
Example 5.2.2. Let f C(U). Then for x U , we have dxf TxU:
dxf() =d
dt
t=0
f , = [] TxU
So, we can define the differential 1-form:
df T U , x 7 dxf
Remark. Note that in our previous notation, df() = f .
In particular, taking the standard coordinates x1, . . . , xn, it is easily seen
that the forms dx1, . . . , dxn are linearly independent at each x U , so theyform a basis for TxU
. We have seen earlier that the derivations 1, . . . , nform a basis for TxU for each x U , and we see now that these are dualbases.
This reveals the meaning behind the notation introduced earlier, and we
see that i, dxi are actually bases for the tangent and cotangent spaces. It
is the natural isomorphism TxRn Rn that leads to the confusion.So, for f C(U), we have the expansion df =ni=1 ifdxi. This defines
an operator d : 0(U) 1(U). We turn now to generalize this notion.
44
-
5.3 The exterior derivative
Definition 5.3.1. We define the exterior derivative, or differential d : k(U)k+1(U) in local coordinates:
d( i1
-
3. First let f C(U). Then
d(df) = d
( ni=1
ifdxi
)=
ni=1
d(if)dxi =
=n
i,j=1
jif dxj dxi
Now, we have dxj dxi = dxi dxj and since f is smooth, jif =ijf . So, the summand above is antisymmetric, and since we are
summing over all i and j, the sum must vanish, so d(df) = 0.
Now let = fdxI . Then d = df dxI . Note that by definition,
d(dxI) = d(1 dxI) = d(1) dxI = 0
Then,
d(d) = d(df dxI) = d(df) dxI + df d(dxI) = 0
As before, by linearity the result follows for a general form.
We again generalize the pullback map:
Definition 5.3.3. For a smooth map f : U V define:
f : k(V ) k(U) , f x(1, . . . , k) = f(x)(f1, . . . , fk)
Proposition 5.3.4.
f () = f f for C(U), k(U). f ( ) = f f . df = f d.
Proof. We prove only the last part. First, for any C(U):
df() = d( f)() = d(f) = f d()
Next, let = dx1 . . . dxk. Then,
f = f f dx1 . . . f dxk = f dfx1 . . . dfxk
46
-
So,
df = df dfx1 . . . dfxk + f d(dfx1) =0
. . . dfxk
. . .+ (1)k1f dfx1 . . . d(dfxk) =0
= df dfx1 . . . dfxk =
= f d f dx1 . . . f dxk = f (d dx1 . . . dxk) = f d
Corollary 5.3.5. The differential is well-defined.
Proof. Let x1, . . . , xn and y1, . . . , yn be two sets of local coordinates, and let
T : M M be the change of coordinates between them:T (x1, . . . , xn) =
(y1(xi), . . . , yn(xi)
)Then T is a diffeomorphism, so for any k(M) we have dT = T d,i.e. its differential is the same given in both coordinate systems.
Example 5.3.6. Consider Rn with a volume form . Let v be a vector fieldon Rn. Then d(iv) n(Rn) and we can define the divergence of v:
div v = v := d(iv)
C(Rn)
For instance, in R3, consider the standard volume form 0 = dx dy dz.Let v = (vx, vy, vz). Then
iv0 = iv(dx dy dz) = ivdx=vx
dy dz dx ivdy=vy
dz + dx dy ivdz=vz
=
= vxdy dz vydx dz + vzdx dyTherefore,
d(iv0) =vxx
dx dy dz vyydy dx dz + vzz
dz dx dy =
=
(vxx
+vyy
+vzz
)dx dy dz
So, div v =vxx
+vyy
+vzz
. This can be generalized - in Rn, we have
div v =ni=1
vixi
.
47
-
5.4 Differential forms on manifolds
Definition 5.4.1. Let M be a smooth manifold. Define
Ek(M) = {(x, ) : x M , k(TxM)}
Ek(M) is called an exterior form bundle on M . Define the bundle projection,
pi : Ek(M)M , pi(x, ) = x
We can define on Ek(M) a smooth structure, similarly to that defined onthe tangent bundle. Thus Ek(M) becomes a smooth manifold.Definition 5.4.2. A smooth differential k-form on M is a smooth map
: M Ek(M) such that pi = 1. The set of smooth k-forms is de-noted k(M).
When k = 1, we denote E1(M) = T M , called the cotangent bundle. For1-forms, we use the notation := ().
Definition 5.4.3. Let k(M). Define the support of :
supp = cl{x M : (x) 6= 0}
The set of k-forms with compact support is denoted kc (M).
We can immediately generalize the notions of the wedge product, differ-
ential and pullback to differential forms on manifolds.
Example 5.4.4. On the cotangent bundle T M of any manifold M , theretwo important forms: Consider bundle projection pi : T M M . For (q, p) T M , we have pi : T(q,p)(T M) TqM . So we can define:
1(T M) , (q,p)() =
Now, let q1, . . . , qn be coordinates on M , and pi := dqi coordinates on TqM.
Then we can write T(q,p)(T M) as
= q1q1 + . . .+ qnqn + p1p1 + . . .+ pnpn := (q, p)
Then pi() = q, so
() == p1q1 + . . .+ p
nqn = p1dq1() + . . .+ pndqn()
48
-
And we see that in local coordinates,
= p1dq1 + . . .+ pndqn
is called the Liouville form. Using the Liouville form, we define the Sym-
plectic form, := d. In local coordinates,
= dp1 dq1 + . . .+ dpn dqn
In particular, we have
n := . . . = n! dp1 dq1 . . . dpn dqn.
We conclude that T M is always orientable, even for non-orientable mani-folds.
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-
6 Integration on manifolds
6.1 Integrals of differential forms
Let Mn be an oriented manifold and nc (M). We wish to define theintegral of over M .
We begin in Rn:
Definition 6.1.1. Let U be an open subset of Rn or the half-space Hn. Let nc (U). Then we have = f dx1 . . . dxn. Define
U
:=
U
fdx1 . . . dxn
Proposition 6.1.2. Let F : U V be a diffeomorphism and nc (V ).Then
U
F = V
where
=
{1 F preserves orientation
1 F reverses orientationProof. Denote 0 = dx
1 . . . dxn and 0 = F 0. Then, as we have seenpreviously,
0(1 . . . , n) = 0(F1, . . . , Fn) = det(F)0(1, . . . , n)
As above, we can write = f 0, with a smooth function f , and we get:
F = F f F 0 = F f det(F)w0
Finally, noting that = sgn det(F) and using the change of variables theo-rem in Rn we get:
U
F =U
F f det(F)0 =U
f F (x) |JF (x)| =det(F)
dx1 . . . dxn =
=
V
f(x)dx1 . . . dxn =
V
We now turn to integration on manifolds:
50
-
Definition 6.1.3. LetMn be an oriented manifold, and {(U, )} a locallyfinite atlas such that : U Rn preserves orientation for all . Let h bea compatible partition of unity. Define:
M
:=
(U)
(1 )(h)
Note that since {U} is locally finite and supp compact, supp U 6= for only finitely many , so this sum is finite.
This definition relies heavily on our choice of atlas and partition of unity,
so we need to prove it is independent of it:
Claim.
M
does not depend on the choice of atlas or partition of unity.
Proof. Let {(V, )} be another atlas and g a compatible partition of unity.Denote by I, J the integrals using the two atlases. Then
I =
(U)
(1 )(h) =
(U)
(1 )g
=(1 )(1)
(1 )(h) =
=,
(U)
(1 )(hg )
Note that supp(h g) W := U V, so we can write
I =,
(W)
(1 )(hg )
By our construction, T := 1 : (W) (W) is an orientation-preserving diffeomorphism. Note that
T(1 )
(gh ) = (1 ) (
1 )
=1
(gh ) = (1 )
(gh )
Then, using proposition 6.1.2 we get(W)
(1 )(hg ) =
(W)
(1 )(hg )
51
-
So,
I =,
(W)
(1 )(hg )
But, retracing our steps above we see that
J =
(V)
(1 )(g) =
,
(W)
(1 )(hg )
Thus I = J and the integral is well-defined.
Proposition 6.1.4.
1.
: nc (M) R is a linear functional.
2.
(M,or)
= (M,or)
.
3. If f : Mn Nn is a diffeomorphism then for any nc (N):M
f = N
where is defined as in proposition 6.1.2.
6.2 Stokes Theorem
LetMn be an oriented manifold. As we have seen, if M 6= it is an orientedmanifold of degree n 1 with the induced orientation. For any differentialform, by convention
= 0
Theorem 6.2.1 (Stokes Theorem). Let n1c (M). ThenM
=
M
d
Proof. By linearity of the integral, it is enough to prove the result for mono-
mials.
First assume M = Rn. Then M = , so we have to showRnd = 0
52
-
Assume WLOG = fdx2 . . .dxn with f Cc (Rn). Then d = 1fdx1. . . dxn. We use Fubinis theorem:
Rnd =
Rn1fdx
1 . . . dxn =
Rn1
dx2 . . . dxn
1fdx1
To see that the last integral vanishes recall that f has compact support, so
in particular for any fixed x2, . . . , xn we have
limx1
f(x1, . . . , xn) = limx1
f(x1, . . . , xn) = 0
So,
1fdx1 = f(, x2, . . . , xn) f(, x2, . . . , xn) = 0 0 = 0
And indeed, M
d = 0
Next, assume M = Hn. Then M = {x1 = 0} Rn1. Here we have toconsider two cases:
(i) = fdx2 . . . dxn. Then:Hnd =
Rn1
dx2 . . . dxn 0
1fdx1 =
=
Rn1
f(0, x2, . . . , xn)dx2 . . . dxn =
Hn
(ii) = fdx1 . . . dxj . . . dxn. WLOG we may assume j = n. Now,as a submanifold of Rn, Hn is defined as the zero set of the smoothfunction x1. Then TxH
n = {dx1}. So Hn
= 0, and then obviouslyHn
= 0
As before, we find thatHnd =
Hn1
dx1 . . . dxn1
nfdxn = 0
53
-
Next, suppose that M is an open subset of Rn or Hn. Then, since hascompact support, we can extend it as zero to all of Rn or Hn, and thus reducethe result to the previous case.
Finally, let Mn be any manifold, and let {(U, )} be a locally finite atlaswith a partition of unity h. Note that {(M U, )} forms an atlas onM , and (U) = (M U). Then, using the previous cases in Rn, weget:
M
d =
(U)
(1 )d(h) =
(U)
d((1 )
(h))=
=
(U)
(1 )(h) =
(MU)
(1 )(h) =
=
M
We now give a few applications os Stokes Theorem, in the form of classical
results from analysis.
Lemma 6.2.2 (Greens formula). Let G R2 be a bounded domain, and let = G. Let f, g C(G). Then
fdx+ gdy =
G
(g
x fy
)dx dy
Proof. Set = fdx + gdy 1(G). Then d =(g
x fy
)dx dy, and
the result follows from Stokes Thorem.
Theorem 6.2.3 (The Divergence Theorem). Let Gn Rn be a submanifoldwith boundary, 0 the standard volume form and v a vector field on G. De-
note by (, ) the Euclidean inner product. Let x be the outward unit normalto G at x. Then
G
(x, v) ix0 =
G
div v 0 =G
ni=1
vixi
0
This integral is called the flux of v through the hypersurface G.
Proof. By definition, TxG = TxG span{x}. Therefore we can writev(x) = w(x) + (x, v)x , w TxG
54
-
Then by linearity of the interior product
iv0 = iw0 + i(x,v)x0 = iw0 + (x, v)ix0
Next, note that iw0G
= 0. Indeed, since w TxG, for any 1, . . . , n1 TxG the vectors w, i must be linearly dependent. Therefore
iw0(1, . . . , n1) = 0(w, 1, . . . , n1) = 0
So,
iv0G
= (x, v)ix0G
Recalling that by definition div v0 = d(iv0) and applying Stokes Theoremwe get:
G
div v 0 =G
iv0 =
G
(x, v)i 0
Example 6.2.4 (The residue formula for analytic functions). Let U C bea domain. For complex functions f : U C we write:
f(z) = f(x, y) = u(x, y) + iv(x, y)
A smooth function f is called analytic is if satisfies the Cauchy-Riemann
equations :u
x=v
y,u
y= v
x
We define complex-valued differential forms as follows. Consider the complex
variable z = x+ iy. We define dz = dx+ idy. Then we define:
f(z)dz = (u+ iv)(dx+ idy) = (udx vdy) + i(vdx+ udy)
Claim. f is analytic iff d(fdz) = 0
Proof. Straightforward computation:
d(udx vdy) = uy
dy dx vx
dx dy = (u
y+v
x
)dx dy
d(vdx+ udy) =v
ydy dx+ u
xdx dy =
(u
x vy
)dx dy
We see that d(fdz) = 0 (C-R).
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We say that f has a pole at a if there is a neighbourhood of a where it
has the representation
f(z) =
k=Nck(z a)k
with N > 0, and the series converges absolutely. The residue of f at a is
res(f, a) = c1.The main result concerning complex integration is:
Theorem 6.2.5 (The residue formula). Let f be analytic in a domain U
except for finitely many poles pj, 1 j n. Let = D be a curve in U ,the boundary of a bounded domain D U which contains all poles. Then
fdz = 2piinj=1
res(f, pj)
Proof. Let Bj be a ball of radius r around pj, and j = Bj, and set V :=
D \nj=1
Bj. As an oriented manifold,
V = (,+) unionsq (1,) unionsq . . . unionsq (n,)Then Stokes Theorem implies:
0 =
V
d(fdz) =
V
fdz =
fdz nj=1
j
fdz
Let us calculate
j
fdz: assume WLOG pj = 0. Then we can find r > 0
such that in a neighbourhood of Bj we have:
f(z) =
k=Nckz
k
On j we have z = reit, therefore zk = rkeikt and dz = i reitdt. So,
j
zkdz = irk+1 2pi0
ei(k+1)tdt
For any k 6= 1 we get:j
zkdz = irk+11
i(k + 1)ei(k+1)t
2pi0
= 0
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For k = 1 we get: j
zkdz = i
2pi0
dt = 2pii
Therefore, we get:j
fdz =
k=Nck
j
zkdz = 2piic1 = 2pii res(f, pj)
So,
fdz =nj=1
j
fdz = 2piinj=1
res(f, pj)
6.3 The Degree Theorem (II)
We return to the degree of a map from a different point of view.
Throughout this section we assume: Mn, Nn are closed oriented manifolds,
with N connected.
Theorem 6.3.1 (2nd Degree Theorem). Let f : M N be a smooth map.Then for any n(N), we have:
M
f = deg(f)N
Remark. Note that in the case where f is a diffeomorphism, we have
deg(f) = 1 = and the theorem reduces to proposition 6.1.2.As with the first Degree Theorem, we begin the proof with several lemmas:
Lemma 1. Suppose Mn = Xn+1. Let F : X N be a smooth map andf = F
M. Then for any n(N),
M
f = 0
Proof. Simply compute:M
f =X
F =X
dF =X
F d
And note that is a top-degree form, so d = 0 and thus the integral
vanishes.
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Lemma 2. Let f0, f1 : M N be homotopic. Then for any n(N),M
f 0 =M
f 1
Proof. Denote X := M [0, 1]. Let F : X N be a homotopy betweenf0, f1. As an oriented manifold,
X = (M {0},) unionsq (M {1},+)
Then using lemma 1 we get:
0 =
X
F = M
f 0 +M
f 1
Lemma 3. Let y be a regular value of f : M N . Then there is a neigh-bourhood y U such that for any n(N) with supp U we have:
M
f = deg(f)N
Proof. Assume f1(y) = {x1, . . . , xk} 6= . We have seen during the proof oflemma 2 of the first degree theorem how to construct neighbourhoods xi Vi, y U such that f
Vi: Vi U is a diffeomorphism and f1(U)
ni
Vi.
Then for any n(N) with supp U , we use proposition 6.1.2:Vi
(fVi) = i
U
And then we get:M
f =ki=1
Vi
f =ki=1
i
U
= deg(f)
N
Next, assume f1(y) = . Then deg(f) = 0. Note that f(M) N is closed,so N \ f(M) is open, and thus we can find a neighbourhood y U such thatU f(M) = . Then for any with supp U , clearly f = 0.Proof of the 2nd Degree Theorem. Let y be a regular value of f and let y Ube a neighbourhood as given by lemma 3. As we have seen before, every z N is isotopic to y, through an isotopy h
(z)t : N N with h(z)1 (y) = z. Then
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{h(z)1 (U)}zN forms an open cover of N , therefore it has a finite subcover,h1(U), . . . , hN(U) where hi := h
(zi)1 . Let 1, . . . , N be a compatible partition
of unity. Set i := i. Then:
(i)
M
f =M
f ( Ni=1
i)=
N1=i
M
f (i) =Ni=1
M
f i
Note that:
hi f is homotopic to f . hi(U) is a neighbourhood as in lemma 3 for hi f around zi. suppi hi(U).
Then:
(i i)
M
f i =M
(hi f) i = deg(hi f)N
i = deg(f)
N
i
Plugging (ii) into (i) we getM
f =Ni=1
deg(f)
N
i = deg(f)
N
( Ni=1
i
) = deg(f)
N
Using this form of the Degree Theorem, we can easily prove the following
important result:
Lemma 6.3.2. Let Lf M g N , with L, M, N closed, and M, N con-
nected. Then deg(g f) = deg(g) deg(f).Proof. Let n(N) with n = dimN . From the second Degree Theoremwe get:
L
(g f) =L
f (g) = deg(f)M
g = deg(f) deg(g)N
The result follows, again, from the second Degree Theorem.
Example 6.3.3. Let M2 R3. Consider the Gauss map:
G : M S2 , p 7 (p)
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In R3 we have the natural isomorphism TpM T(p)S2, so we may consider:G : TpM TpM
Define the Gauss curvature of M , (p) := detG(p).Let = dx dy dz be the standard volume form. Define the inducedvolume forms, := in on S
2 and := i on M .
Claim. G = .
Proof. Note that by our isomorphism TaM T(a)S2, . Then theclaim is simply the definition of the determinant.
Then, using the Degree Theorem we get:M
=
M
G = deg(G)S2 = 4pi deg(G)
Recall that we have seen (theorem 4.3.13) that for a hypersurface M2k R2k+1, its Euler characteristic is given by (M) = 2 deg(G) . Moreover, for
a surface of genus g, we have seen that (M) = 2 2g. So, deg(G) = 1 g.We get the following famous result:
Theorem 6.3.4 (Gauss-Bonnet Theorem). Let M2 R3 be a surface ofgenus g. Then:
M
= 4pi(1 g)
6.4 Linking number and the Gauss integral
Let 1, 2 : S1 R3 be two closed, disjoint curves. We are concerned with
the question - can they be pulled apart?
More formally, we are looking for a homotopy t1, t2 which satisfies:
1. 01 = 1 , 02 = 2.
2. t1 t2 = for all t.3. 11 R3+ , 12 R3.We begin by defining a homotopy invariant:
Definition 6.4.1. Let 1 , 2 be two curves as before. Set:
f : S1 S1 S2 , (s, t) 7 1(s) 2(t)||1(s) 2(t)||Define the linking number of 1, 2 as lk(1, 2) = deg(f).
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Note that lk(1, 2) is indeed invariant with respect to a homotopy satis-
fying conditions 1, 2 above.
Example 6.4.2. Suppose 1, 2 satisfy conditions 13 above. Then we canfind a homotopy such that dist(1, 2) is arbitrarily large. Then 1(s)2(t)varies very little with t, s, and so f is not onto. Thus lk(1, 2) = 0.
Example 6.4.3. Let 1 be a circle or radius 1 in the x, y plane, and 2 a
curve in the x, z plane through the origin O, oriented as in figure 4:
Figure 4: Linking number
Let P = (1, 0, 0). Then f(P,O) = (1, 0, 0), and clearly, there areno other points P , Q such that f(P , O) = (1, 0, 0). Thus it remains tocompute sgn detD(P,Q)f . As this is a local computation of the first order, we
may approximate 1, 2 by straight lines:
1(s) = (1,s, 0) , 2(t) = (0, 0, t)
Then 1(s) 2(t) = (1,s,t), so:
f(s, t) =1
1 + s2 + t2(1,s,t) =
= (1,s,t) + o(||(s, t)||)
Note that at f(0, 0) = (1, 0, 0) S2, the unit normal is (1, 0, 0) = x.Then the induced orientation on S2 is dy dz = dz dy. Thus we take
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the orientation preserving coordinate chart (z, y). Then f has the local
representation:
f : (s, t) 7 (t,s) + o(||(s, t)||)Then,
D0f =
(0 11 0
) det(D0f) = 1
So, lk(1, 2) = deg(f) = 1.Definition 6.4.4. Define the vector product :
[, ] : R3 R3 R3 , [a, b] =x y za1 a2 a3b1 b2 b3
Clearly, the vector product is bilinear and skew-symmetric. Moreover, we
have:
Proposition 6.4.5. For any a, b, c R3:
(a, [b, c]) = dx1 dx2 dx3(a, b, c) =a1 a2 a3b1 b2 b3c1 c2 c3
Corollary 6.4.6. [a, b] a, b.Proof. Using the previous proposition we get (a, [a, b]) = (b, [a, b]) = 0.
Lemma 6.4.7 (The Gauss Formula).
lk(1, 2) =1
4pi
T2
(1(s) 2(t), [2(t), 1(s)]
)||1(s) 2(t)||3 ds dt
Proof. Let be the standard volume form on S2. Set
f : (s, t) 7 1(s) 2(t)||1(s) 2(t)||From the Degree Theorem we get:
T2f = deg(f)
S2 = 4pi deg(f)
Or,
(i) lk(1, 2) = deg(f) =1
4pi
T2f
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Let x S2. Then the normal at x is nx = x, so for any , TxS2, usingproposition 6.4.5 we get:
(, ) = dx1 dx2 dx3(x, , ) = (x, [, ])Then, for , TzT2 we have
f (, ) = (f, f) =(f(z), [f, f]
)Take coordinates s, t on T2 = S1 S1. Then ds dt is a volume form, so wecan write f = A(s, t)ds dt. We have:
A(s, t) = f (s, t) =(f(s, t), [fs, ft]
)=(f(s, t), [sf, tf ]
)Let us perform a general computation. Suppose v : R R3 \ {0}. Compute:
d
dt
(v(t)
||v(t)||)=
v(t)
||v(t)|| +d
dt
(1
||v(t)||)v(t)
Note that ||v(t)|| =(
v(t), v(t)). So,
d
dt
(1
||v(t|)|)= 1||v(t)||2
1
2(
v(t), v(t)) 2(v(t), v(t)) = (v(t), v(t))||v(t)||3
Combining these computations, we get:
d
dt
(v(t)
||v(t)||)=
v(t)
||v(t)|| (v(t), v(t)
)||v(t)||3 v(t)
Setting v(s, t) = 1(s) 2(t) yields:
sf =sv
||v|| (v, sv
)||v||3 v , tf =
tv
||v|| (v, tv
)||v||3 v
[sf, tf ] = [sv, tv]||v||2 + [, v]
For some vector . Note that sv = 1(s), tv = 2(t). Therefore,
A(s, t) =(f, [sf, tf ]
)=
(v
||v|| ,[sv, tv]
||v||2 + [, v])=
=1
||v||3(v, [1(s),2(t)]
)+
1
||v||(v, [, v]
) =0
=
=
(1(s) 2(t), [2(t), 1(s)]
)||1(s) 2(t)||3
Plugging this into (i), the result follows.
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7 The Lie derivative
7.1 The Lie derivative
Let M be a manifold, and a vector field on M with compact support. We
consider the flow of , i.e. the solution to the ODE
dg
dt(t, x) =
(g(t, x)
)Set gt(x) := g(t, x). Then gt : M M defines a smooth family of diffeomor-phisms of M , and g0 = 1.
Let k(M). Set t := gt. Then t represents the time evolution of under the flow gt.
Definition 7.1.1. The Lie derivative of (along ) is:
L =d
dt
t=0
t
Example 7.1.2. Let f 0(M) = C(M). Then
Lf =d
dt
t=0
gt f =d
dt
t=0
f gt = df() = f
Example 7.1.3. Let k(M) such that L = 0.Claim. t = g
t =
Proof. We use the fact, known from the theory of ODE, that for the flow of
a vector field, gt+s = gt gs. Then t+s = gts, so,d
dtt = lim
s0wt+s wt
s= lim
s0gtws gtw0
s=
= gt lims0
ws w0s
= gtL = 0
So, t = 0 = .
Proposition 7.1.4. The Lie derivatives satisfies:
1. Linearity: L(a + b ) = aL+ bL.2. Leibnitz rule: L( ) = L + L.3. Ld = dL.
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-
4. For f C(M), Lf = f .Proof.
1. Obvious.
2. ( )t = gt ( ) = gt gt = t t. Then the result followsfrom the Leibnitz rule for normal derivatives.
3. Note that (d)t = gt d = dg
t = dt. So,
Ld =d
dt
t=0
(d)t =d
dt
t=0
dt = dd
dt
t=0
t = dL
To see that the third equality holds, note that written in local coordi-
nates it simply means that the time derivation commutes with partial
derivatives.
4. See example 7.1.2 above.
Corollary 7.1.5 (Cartans formula). L = di + id.
Proof. Denote L = di + id. Then straightforward inspection shows that L
satisfies 1-4 above. SetD = LL. ThenD satisfies 1-3, and for f C(M),Df = 0. Then also Ddf = dDf = 0. Let = fdx1 . . . dxk k(M).Then,
D = Df=0
dx1 . . . dxk + f Ddx1 =0
. . . dxk + . . .+ fdx1 . . . Ddxk =0
= 0
From linearity, it follows that D = 0 for any k-form. So, D 0 and henceL = L.
7.2 Application: Hamiltonian mechanics
Let V n be a linear space, and 2(V ). Define
I : E E , 7 i
Excercise 7.2.1.
1. If I is an isomorphism, then n is even.
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2. Let n = 2k. Then I is an isomorphism iff k 6= 0.
Definition 7.2.2. Let M2n be a manifold. A symplectic form on M