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Math 240A: Differentiable Manifolds andRiemannian Geometry
Simon Rubinstein–Salzedo
Fall 2005
0.1 Introduction
These notes are based on a graduate course on differentiable manifolds and Rieman-nian geometry I took from Professor Doug Moore in the Fall of 2005. The text-books were An Introduction to Differentiable Manifolds and Riemannian Geometryby William Boothby and Calculus on Manifolds by Michael Spivak. Many otherbooks are also mentioned in the notes. Since the professor handed out very goodnotes, I have made very few changes to these notes.
1
Chapter 1
September 22, 2005
Calculus of several variables starts with calculus on Rn and then proceeds to “calculuson manifolds.”
Let U be an open subset of Rn and f : U → R a function. f is C0 iff f is continuous.f is Ck iff f has continuous partial derivatives of order at most k. f is C∞ iff fhas continuous partial derivatives of all orders. f is Cω iff f can be extended in aconvergent Taylor series expansion about any point.
Suppose F : U → Rn, F = (f 1, . . . , fn). F is C0, Ck, C∞, or Cω iff each f i is.
Let M be a topological space. An n-dimensional chart or coordinate system onM is a pair (U,ϕ) such that U is an open subset of M and ϕ is a homeomorphismfrom U onto an open subset ϕ(U) of Rn.
Let r ∈ 0, 1, 2, . . . , k, . . . ,∞, ω. An n-dimensional Cr-atlas on M is a collectionA = (Uα, ϕα) : α ∈ A of n-dimensional charts such that
1. M =⋃Uα : α ∈ A.
2. ϕβ ϕ−1α is a Cr map where defined for all α, β ∈ A.
Two n-dimensional Cr atlases A1 and A2 are equivalent if A1 ∪ A2 is a Cr-atlas.
2
Definition. An n-dimensional Cr-manifold is a Hausdorff second countable topo-logical space M together with an equivalence class of Cr-atlases.
Examples.
1. Let U be an open subset of Rn. Then (U, id) is a Cr-atlas making U into ann-dimensional Cr-manifold, for any r.
2. Let Sn = (x1, . . . , xn+1) ∈ Rn+1 : (x1)2 + · · · + (xn+1)2 = 1. Let n =(0, . . . , 0, 1), s = (0, . . . , 0,−1), U = Sn − n, and V = Sn − s. Defineϕ : U → Rn by
ϕ(x1, . . . , xn+1) =1
1− xn+1(x1, . . . , xn) = (u1, . . . , un).
Then (u1, . . . , un, 0) = (0, . . . , 0, 1) + λ(x1, . . . , xn+1 − 1), where λ = 11−xn+1 . ϕ
is a homeomorphism from U to Rn because we can solve for (x1, . . . , xn+1) interms of (u1, . . . , un):
(u1)2 + · · ·+ (un)2 =(x1)2 + · · ·+ (xn)2
(1− xn+1)2=
1− (xn+1)2
(1− xn+1)2=
1 + xn+1
1− xn+1.
Thus |u|2(1−xn+1) = 1+xn+1, so xn+1(1+|u|2) = |u|2−1. Hence xn+1 = |u|2−1|u|2+1
,
and 1− xn+1 = 2|u|2+1
,
(x1, . . . , xn) = (1− xn+1)(u1, . . . , un) =2
|u|2 + 1(u1, . . . , un)
(x1, . . . , xn+1) = ϕ−1(u1, . . . , un) =1
|u|2 + 1(2u1, . . . , 2un, |u|2 − 1).
Similarly, we define ψ : V → Rn by
ψ(x1, . . . , xn+1) =1
1 + xn+1(x1, . . . , xn+1) = (v1, . . . , vn).
We claim thatA = (U,ϕ), (V, ψ) is a Cr atlas making Sn into an n-dimensionalCr-manifold. Indeed,
ψ ϕ−1(u1, . . . , un) =1− xn+1
1 + xn+1(u1, . . . , un) =
1
|u|2(u1, . . . , un),
3
so ψ ϕ−1 : Rn − 0 → Rn is Cr. Similarly,
ϕ ψ−1(v1, . . . , vn) =1
|v|2(v1, . . . , vn).
3. Suppose Mm and Nn are Cr manifolds of dimensions m and n with Cr atlases(Uα, ϕα) : α ∈ A and (Vβ, ψβ) : β ∈ B respectively. Then Mm×Nn is a Cr
manifold of dimension m+n with Cr atlas (Uα×Vβ, ϕα×ψβ) : α ∈ A, β ∈ B.
Suppose Mm and Nn are Cr manifolds with Cr atlases (Uα, ϕα) : α ∈ A and(Vβ, ψβ) : β ∈ B. Then a continuous map F : Mm → Nn is a Cr map if ψβF ϕ−1
α
is Cr where defined, for α ∈ A and β ∈ B.
Clearly the composition of two Cr maps is a Cr map. A Cr map F : Mm → Nn isa Cr diffeomorphism if there is a Cr map G : Nn → Mm such that G F is theidentity on M and F G is the identity on N . Two Cr manifolds M and N are Cr
diffomorphic if there is a Cr diffeomorphism F : M → N .
Basic Problem. Classify n-dimensional Cr manifolds up to Cr diffeomorphism.
Any C1 manifold is C1 diffomorphic to a unique Cr manifold for any r ≥ 1.
On the other hand Milnor (1956) showed that there are C∞ manifolds which arehomeomorphic but not diffeomorphic to S7. Kervaire (1960) showed that there areC0 manifolds of dimension 10 that cannot be made into C1 manifolds.
Terminology. A C0 manifold is called a topological manifold. A C∞ manifold iscalled a smooth manifold.
n = 1: Any topological one-manifold has a unique smooth structure. The only con-nected one-dimensional manifolds are R and S1. (This is proven in Milnor, Topologyfrom the Differentiable Viewpoint, Appendix.)
4
n = 2: Any topological two manifold has a unique smooth structure. A two-dimensional manifold is said to be orientable if it has an atlas (Uα, ϕα) : α ∈ Asuch that ϕβϕ−1
α is orientation-preserving where defined. (A precise definition will begiven later.) It can be proven that any compact connected orientable two-dimensionalsmooth manifold is diffeomorphic to Σg, a sphere with g handles. (This is proven, forexample, in Andrew Wallace, Differential Topology: First Steps, Chapter 7.)
n = 3: Any topological 3-manifold has a unique smooth structure. A 3-manifold Mis said to be simply connected if any continuous map
f : S1 = z ∈ C : |z| = 1 →M
can be extended to a continuous map
F : D2 = z ∈ C : |z| ≤ 1 →M.
Poincare Conjecture. Any compact simply connected 3-manifold is homeomorphicto S3.
Posed as a question by Poincare in 1904, this is one of the 7 “millenium prize prob-lems” posed by the Clay Mathematics Institute. (See http://www.claymath.org.)
Perelman claims a proof based upon “Ricci flow.” The proof involves nonlinear PDEsand differential geometry.
n = 4: R4 has infinitely many distint smooth structures. Some compact four-dimensional topological manifolds have more than one smooth structure; others havenone.
This follows from work of Freedman (1982) and Donaldson (1983). Donaldson’s workis based upon the nonlinear PDEs from Yang-Mills theory and reverses the usual
5
direction of flow of applications — in Donaldson’s work,
(physics) → (nonlinear PDEs) → (topology and geometry).
6
Chapter 2
September 27, 2005
Another example of a manifold. Let Mn = P n(R) be the set of lines through theorigin in Rn+1. Define an equivalence relation ∼ on Rn+1 − 0 by (x1, . . . , xn+1) ∼(y1, . . . , yn+1) iff there exists a λ ∈ R−0 such that (x1, . . . , xn+1) = λ(y1, . . . , yn+1).P n(R) is the space of equivalence classes. Let [x1, . . . , xn+1] denote the equivalenceclass of (x1, . . . , xn+1) and define π : Rn+1 − 0 → P n(R) by π(x1, . . . , xn+1) =[x1, . . . , xn+1].
We give P n(R) the quotient topology. Thus U ⊆ P n(R) is open iff π−1(U) is open inRn+1 − 0.
With this topology, P n(R) is Hausdorff and second countable. Let Ui = [x1, . . . , xn+1] ∈P n(R) : xi 6= 0, for 1 ≤ i ≤ n + 1. Ui is the set of lines in Rn+1 that do not lie inthe hyperplane xi = 0.
Define ϕ1 : U1 → Rn by ϕ1([x1, . . . , xn+1]) =
(x2
x1 , . . . ,xn+1
x1
), ϕ2 : U2 → Rn by
ϕ2([x1, x2, . . . , xn+1]) =
(x1
x2 ,x3
x2 , . . . ,xn+1
x2
), and so forth. We claim that (Ui, ϕi) :
1 ≤ i ≤ n+ 1 is a C∞ (or smooth) atlas on P n(R).
7
Indeed, if i < j,
ϕj ϕ−1i (u1, . . . , un) = ϕj([u
1, . . . , ui−1, 1, ui, . . . , un])
=
(u1
uj−1, . . . ,
ui−1
uj−1,
1
uj−1,ui+1
uj−1, . . . ,
uj−2
uj−1,uj
uj−1, . . . ,
un
uj−1
),
and this map is C∞ on
ϕi(Ui ∩ Uj) = (u1, . . . , un) ∈ Rn : uj−1 6= 0.
There is a similar formula for ϕi ϕ−1j .
Thus P n(R) is a smooth manifold of dimension n.
Many more manifolds will be constructed painlessly once we have the Inverse Func-tion Theorem.
2.1 Calculus of Several Variables in Rn
Let U be an open subset of Rn, F : U → Rm a continuous map.
F is said to be differentiable at x0 ∈ U if there is a linear map T : Rn → Rm suchthat given ε > 0, there exists δ > 0 such that
‖h‖ < δ ⇒ ‖F (x0 + h)− F (x)− T (h)‖‖h‖
< ε.
We can also write this as follows: F is differentiable at x0 if there is a linear mapT : Rn → Rm such that
F (x0 + h) = F (x0) + T (h) + Error(h),
where the error term Error(h) satisfies
lim‖h‖→0
‖Error(h)‖‖h‖
= 0.
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If F is differentiable at x0, the affine mapping L : Rn → Rm defined by
L(x) = F (x0) +DF (x0)(x− x0)
is called the linearization of F at x0.
In terms of the standard coordinates (x1, . . . , xn) on Rn, DF (x0) is represented bythe matrix
DF (x0) =
∂f1
∂x1 (x0) · · · ∂f1
∂xn (x0). . .
∂fm
∂x1 (x0) · · · ∂fm
∂xn (x0)
.
The component ∂f i
∂xj (x0) of this matrix is called the partial derivative of f i withrespect to xj.
If F : U → Rm is differentiable at every x ∈ U , we have a map
DF : U → L(Rn,Rm) = linear maps from Rn to Rm.
We say that F is C1 if DF : U → L(Rn,Rm) is continuous.
L(Rn,Rm) can be regarded as the Euclidean space Rnm. Hence we ask whether DFis differentiable. We say that F is C2 if DF is C1. Inductively, F is Ck if DF isCk−1. F is C∞ if it is Ck for all k.
If F is C2, then for all x ∈ U ,
D2F (x) = D(DF )(x)
∈ L(Rn, L(Rn,Rm))
= L2(Rn,Rm)
= bilinear maps φ : Rn × Rn → Rm.
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More generally, if F is Ck, then
DkF (x) ∈ Lk(Rn,Rm) =
φ : Rn × · · · × Rn︸ ︷︷ ︸k
→ Rm : φ is multilinear
.
Proposition. (Chain Rule.) If U and V are open subsets of Rn and Rp, and F :U → V and G : V → Rm are C1, then G F : U → Rm is C1, and
D(G F )(x) = DG(F (x))DF (x), for x ∈ U. (∗)
Sketch of Proof. If x0 ∈ U , F (x0+h) = F (x0)+DF (x0)(h)+ErrorF (h), G(F (x0)+k) = G(F (x0)) +DG(F (x0))k + ErrorG(k), and hence
(G F )(x0 + h) = G(F (x0)) +DG(F (x0))(DF (x0)h+ ErrorF (h))
+ ErrorG(DF (x0)h+ ErrorF (h)),
so we need only verify that
lim‖h‖→0
‖DG(F (x0))(ErrorF (h))‖‖h‖
= 0
and
lim‖h‖→0
‖ErrorG(DF (x0)h+ ErrorF (h))‖‖h‖
=
lim‖h‖→0
‖ErrorG(DF (x0)h+ ErrorF (h))‖‖DF (x0)h+ ErrorF (h)‖
· ‖DF (x0)h+ ErrorF (h)‖‖h‖
.
In the last expression, the first term goes to zero, and the second term is bounded.The verifications are relatively straightforward. Reference: Boothby, page 27.
Complement. Using (∗) and the Leibniz rule, we can show that if F and G are Ck,then G F is Ck.
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Special Case. Suppose γ : [a, b] → Rn is a smooth curve and F : Rn → Rm is asmooth map. We can write
γ(t) =
x1(t)...
xn(t)
and Dγ(t) = γ′(t) =
dx1
dt(t)...
dxn
dt(t)
.
Think of γ(t) as the trajectory of a moving particle and γ′(t) as its velocity. Then
D(F γ)(t) = DF (γ(t))Dγ(t) ⇒ (F γ)′(t) = DF (γ(t)) · γ′(t).
Thus DF takes the velocity vector to γ to the velocity vector to F γ. We willencounter this fact again later.
Inverse Function Theorem. If U1 and U2 are open subsets of Rn with x0 ∈ U1 andF : U1 → U2 is a C∞ map such that DF (x0) ∈ L(Rn,Rn) is invertible, then there areopen neighborhoods V1 of x0 and V2 of f(x0) with V1 ⊆ U1 and V2 ⊆ U2 and a C∞
map G : V2 → V1 such that
G F = idV1 and F G = idV2 .
Moreover, DG(F (x)) = [DF (x)]−1 for x ∈ V1.
Strategy of Proof. Given y near F (x0), we want to solve F (x) = y for x. To dothis we need only find a fixed point to the map Gy defined by Gy(x) = x+ y − F (x)and apply:
Contraction Map Lemma. Let (X, d) be a complete metric space, H : X → X amap such that there is a constant λ, 0 < λ < 1, such that
d(H(x), H(y)) ≤ λd(x, y) for x, y ∈ X.
Then H has a unique fixed point.
Proof. See Boothby, pages 42 and 43.
11
Chapter 3
September 29, 2005
Inverse Function Theorem. Let U be an open neighborhood of x0 in Rn, F : U →Rn a C∞ map such that DF (x0) is invertible. Then there exist open neighborhoodsV1 of x0 and V2 of F (x0) and a C∞ map G : V2 → V1 such that
G F = idV1 and F G = idV2 .
Moreover, DG(F (x0)) = (DF (x0))−1.
Proof. Assume without loss of generality that x0 = F (x0) = 0 and DF (x0) = I. Wewant to solve F (x) = y by finding a fixed point for Hy(x) = x+ y− F (x). Note thatDHy(0) = I − I = 0. Therefore there exists δ > 0 such that if ‖x‖ ≤ δ, then x ∈ Uand ‖DHy(x)‖ < 1
2.
Let V2 = B δ2(0) =
y ∈ Rn : ‖y‖ < 1
2
. For y ∈ V2, define Hy : Bδ(0) → Rn by
Hy(x) = x+ y − F (x).
12
If x, x′ ∈ Bδ(0), then
‖Hy(x)−Hy(x′)‖ =
∥∥∥∥∫ 1
0
d
dt(Hy(tx+ (1− t)x′)) dt
∥∥∥∥=
∥∥∥∥∫ 1
0
DHy(tx+ (1− t)x′)(x− x′) dt
∥∥∥∥≤∫ 1
0
‖DHy(tx+ (1− t)x′)‖ ‖x− x′‖ dt
≤ 1
2‖x− x′‖.
Note that Hy(0) = y, so ‖Hy(0)‖ < δ2. Hence ‖Hy(0)−Hy(x)‖ ≤ δ, so Hy : Bδ(0) →
Bδ(0). Moreover, ‖Hy(x)−Hy(x′)‖ ≤ 1
2‖x− x′‖ implies that Hy is a contraction, so
by the Contraction Map Lemma, Hy has a unique fixed point.
Define G : V1 → U by G(y) = unique fixed point of Hy. Let V1 = G(V2). Then
G F = idV1 and F G = idV2 .
Claim: G is continuous. Indeed,
‖x− x′‖ ≤ ‖(x− F (x))− (x′ − F (x′))‖+ ‖F (x)− F (x′)‖≤ ‖Hy(x)−Hy(x
′)‖+ ‖F (x)− F (x′)‖
≤ 1
2‖x− x′‖+ ‖F (x)− F (x′)‖,
and hence 12‖x− x′‖ ≤ ‖F (x)− F (x′)‖, which implies that
‖G(y)−G(y′)‖ ≤ 2‖y − y′‖. (∗)
Claim: G is differentiable at each y ∈ V2. Suppose x = G(y), x′ = G(y′). Then
y′ − y = F (x′)− F (x) = DF (x)(x′ − x) + Errorx(x′ − x),
13
where limx′→x‖Errorx(x′−x)‖
‖x′−x‖ = 0.
G(y′)−G(y) = x′ − x = (DF (x))−1(y′ − y) + Errorx(x′ − x).
It follows from (∗) that limy′→y‖Errorx(x′−x)‖
‖y′−y‖ = 0. Thus G is differentiable at y and
DG(y) = (DF (G(y)))−1 or DG(F (x)) = (DF (x))−1.
Since F is C∞ and DG = (DF G)−1, G is C1.
Inductively, we see that since G is Ck, DF G is Ck, so DG = (DF G)−1 is Ck, soG is Ck+1. Hence G is C∞, and the inverse function theorem is proven.
Corollary 1. Let U be an open neighborhood of 0 in Rm, F : U → Rn a smoothmap such that F (0) = 0. If DF (0) is injective, there is a diffeomorphism G from oneopen neighborhood of 0 onto another such that
G F (x1, . . . , xm) = (x1, . . . , xm, 0, . . . , 0).
Proof. Let F =
f1
...fn
, where each f i is real-valued. We can assume without loss
of generality that ∂f1
∂x1 (0) · · · ∂f1
∂xm (0)...
...∂fm
∂x1 (0) · · · ∂fm
∂xm (0)
is nonsingular. Define F : U × Rn−m → Rn by
F (x1, . . . , xn) = F (x1, . . . , xm) +
0...0
xm+1
...xn
.
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Then ∂f1
∂x1 (0) · · · ∂f1
∂xm (0)...
... 0∂fm
∂x1 (0) · · · ∂fm
∂xm (0)∗ I
,
and hence DF (0) is invertible. By the inverse function theorem, F possesses a localinverse G, and
G F (x1, . . . , xm) = G F (x1, . . . , xm, 0, . . . , 0)
= (x1, . . . , xm, 0, . . . , 0),
as desired.
Corollary 2. Suppose U is an open subset of Rn, 0 ∈ U , and F : U → Rm is a smoothfunction such that F (0) = 0. If DF (0) is surjective, there is a local diffeomorphismG from some neighborhood of 0 in Rn onto another such that
F G(x1, . . . , xn) = (x1, . . . , xm).
Proof. Let F =
f 1
...fm
and assume without loss of generality that
∂f1
∂x1 (0) · · · ∂f1
∂xm (0)...
...∂fm
∂x1 (0) · · · ∂fm
∂xm (0)
is nonsingular. Define F : U → Rn by
F (x1, . . . , xn) =
f 1(x1, . . . , xn)...
fm(x1, . . . , xn)xm+1
...xn
,
15
so that if π : Rn → Rm is the projection, π F = F . Then
DF (0) =
∂f1
∂x1 (0) · · · ∂f1
∂xm (0)...
... ∗∂fm
∂x1 (0) · · · ∂fm
∂xm (0)0 I
is nonsingular, so by the inverse function theorem, F has a local inverse G. But then
F G(x1, . . . , xn) = π F G(x1, . . . , xn)
= π(x1, . . . , xn)
= (x1, . . . , xm),
as desired.
16
Chapter 4
October 4, 2005
Let M be an n-dimensional Cr-manifold with Cr atlas A = (Uα, ϕα) : α ∈ A. ACr-chart on M is an n-dimensional chart (U,ϕ) on M such that ϕϕ−1
α and ϕα ϕ−1
are Cr where defined, for α ∈ A.
A Cr map F : M → N between Cr manifolds is a continuous map such that ψF ϕ−1
is Cr where defined, whenever (U,ϕ) is a Cr chart on M and (V, ψ) is a Cr chart onN .
Definition. A Cr map F : M → N with r ≥ 1 is an immersion if D(ψ F ϕ−1)(ϕ(p)) is injective and a submersion if D(ψ F ϕ−1)(ϕ(p)) is surjective forp ∈M , where (U,ϕ) and (V, ψ) are Cr charts on M and N with p ∈ U and F (p) ∈ V .
Definition. A Cr map, r ≥ 1, is an imbedding if it is a one-to-one immersionand F maps M homeomorphically onto F (M), where F (M) is given the subspacetopology it inherits from N .
Examples.
1. γ : R → R2, γ(t) = (t3, t2), is not an immersion because γ′(0) = 0. Note thatthe image of γ has a cusp at the point where it fails to be an immersion.
2. γ : R → R2, γ(t) = (cos t, sin t), has the property that γ′(t) = (− sin t, cos t) isnever zero, so γ is an immersion.
17
3. γ : (−π, π) : R2, γ(t) =(sin t, 1
2sin 2t
), is a one-to-one immersion, but not an
imbedding.
4. The inclusion map i : Sn → Rn+1 is an imbedding, when Sn is given the smoothstructure described in Lecture 1.
Definition. A subset M of an n-dimensional Cr manifold N is said to be an m-dimensional Cr submanifold of N if whenever p ∈M , there is a Cr chart (U,ϕ) onN with p ∈ U such that
ϕ(M ∩ U) = (Rm × (0, . . . , 0)) ∩ ϕ(U). (∗)
We claim that if M is an m-dimensional Cr submanifold of N , then it is also anm-dimensional Cr manifold, and the inclusion i : M → N is a Cr imbedding. Indeed,M is Hausdorff and second countable in the subspace topology.
If p ∈M , let (Up, ϕp) be a Cr chart on N satisfying (∗) with p ∈ Up. Let
π : Rn = Rm ⊕ Rn−m → Rm
be the orthogonal projection. Then (Up ∩M,π ϕp) : p ∈ M is a Cr atlas on Mbecause
(π ϕp) (π ϕq)−1 = π (ϕp ϕ−1q ) |ϕq(Up∩Uq)∩(Rm×0)
is Cr. Moreover, i : M → N is clearly Cr.
Proposition 1. If F : Mm → Nn is a Cr imbedding, then F (M) is a Cr submanifoldof N and F : M → F (M) is a diffeomorphism.
Proof. This is a consequence of Corollary 1 to the Inverse Function Theorem. Sup-pose p ∈ M , q = F (p) ∈ N . We need to construct a Cr chart (W,σ) with q ∈ Wsuch that
σ(W ∩ F (M)) = σ(W ) ∩ (Rm × 0). (∗)
18
Let (U,ϕ) and (V, ψ) be Cr charts on M and N such that p ∈ U , q ∈ V , ϕ(p) =0 = ψ(q). Since F is an immersion, D(ψ F ϕ−1)(0) has rank m = dimM . ByCorollary 1 of the Inverse Function Theorem, there is a diffeomorphism G defined onsome open neighborhood of 0 in Rn such that
G ψ F ϕ−1(x1, . . . , xm) = (x1, . . . , xm, 0, . . . , 0),
for (x1, . . . , xm) in some open neighborhood ϕ(U ′) of 0. Since F maps M homeomor-phically into F (M), if W is a sufficiently small neighborhood of q, F−1(W ) ⊆ U ′.Then (W,σ) = (W,G ψ |W ) is a Cr chart satisfying (∗) with q ∈ W .
Definition. A point q ∈ N is a regular value for the Cr map F : M → N ifwhenever p ∈ F−1(q), D(ψ F ϕ−1)(ϕ(p)) is surjective, whenever (U,ϕ) and (V, ψ)are Cr charts on M and N with p ∈ U , q ∈ V .
Thus if F : M → N is a submersion, every q ∈ N is a regular value.
Proposition 2. If F : Mm → Nn is a Cr map and q ∈ N is a regular value for F ,then F−1(q) is a Cr submanifold of M .
Proof. This is a consequence of Corollary of the Inverse Function Theorem. Supposep ∈ F−1(q). Let (U,ϕ) and (V, ψ) be Cr charts on M and N with p ∈ U , q ∈ V ,ϕ(p) = 0 = ψ(q). Since q is a regular value, D(ψ F ϕ−1)(0) has rank m = dimN .By Corollary 2 of the Inverse Function Theorem, there is a diffeomorphism G on someopen neighborhood of 0 in Rn such that
(ψ F ϕ−1 G)(x1, . . . , xn) = (x1, . . . , xm).
Define H : Rn → Rn by H(x1, . . . , xn) = (xm+1, . . . , xn, x1, . . . , xm). If W is a suf-ficiently small neighborhood of p in M , then (W,σ) = (W,H G−1 ϕ |W ) is a Cr
chart such that σ(W ∩ F−1(q)) = (Rn−m × 0) ∩ σ(W ).
Examples. Define F : Rn+1 → R by
F (x1, . . . , xn+1) = (x1)2 + · · ·+ (xn+1)2.
19
Then q ∈ R is a regular value iff q 6= 0. If q < 0, F−1(q) = ∅, a manifold by default.(The empty set is a manifold of any dimension!) If q > 0, F−1(q) is a sphere of radius√q. Thus Proposition 2 gives a Cr manifold structure on Sn = F−1(1) without having
to construct explicit charts.
20
Chapter 5
October 6, 2005
Let M(n,R) = n×n real matrices, which we can regard as a smooth manifold, dif-feomorphic to Rn2
. Let GL(n,R) = A ∈M(n,R) : detA 6= 0, an open submanifoldof Rn2
.
If B ∈ GL(n,R), define LB : GL(n,R) → GL(n,R) by LB(A) = B · A. LB is a dif-feomorphism with inverse LB−1 . Finally, let Sym(n,R) = A ∈ M(n,R) : AT = A.Define F : GL(n,R) → Sym(n,R) by F (A) = ATA.
Claim: I is a regular value for F and hence
O(n) = F−1(I) = A ∈ GL(n,R) : ATA = I
is a smooth submanifold of GL(n,R).
To see this, note first that DF (A)B = ATB + BTA. If A = I, then DF (I)B =B +BT = 2(symmetric part of B). Note that if B ∈ O(n),
F (LB(A)) = F (BA)
= (BA)T (BA)
= ATBTBA
= ATA
= F (A),
so F LB = F . Then D(F LB)(I) = DF (I) is surjective, and hence DF (B)DLB(I)is surjective, which implies that DF (B) is surjective. Thus I is indeed a regular value.
21
5.1 Tangent Space
Let M be a smooth manifold, p ∈ M . Let F (p) be the set of smooth real-valuedfunctions defined on some open neighborhood of p in M .
Definition. A tangent vector to M at p is a map v : F (p) → R such that
1. If f = g in some neighborhood of p, then v(f) = v(g).
2. v is R-linear: v(cf + g) = cv(f) + v(g), for c ∈ R, f, g ∈ F (p).
3. v satisfies the Leibniz rule: v(fg) = f(p)v(g) + g(p)v(f), for f, g ∈ F (p).
Let TpM = tangent vectors to M at p. TpM is a real vector space, called thetangent space to M at p. For example, suppose M = Rn and v = (a1, . . . , an). Wecan define a corresponding map v : F (p) → R by
v(f) =n∑i=1
ai∂f
∂xi(p).
Then one readily verifies that v ∈ TpRn.
More generally, suppose that M is an n-dimensional manifold, p ∈M , (U,ϕ) a chartwith p ∈ U , ϕ = (x1, . . . , xn). We define ∂
∂xi
∣∣p∈ TpM by
∂
∂xi
∣∣∣∣p
(f) =∂
∂xi(f ϕ−1)(ϕ(p)),
where ∂∂xi is the ith partial derivative. If (a1, . . . , an) ∈ Rn, then
v =n∑i=1
ai∂
∂xi
∣∣∣∣p
∈ Rn.
We say that (a1, . . . , an) are the components of v with respect to the coordinate sys-tem (x1, . . . , xn).
22
Note that∂
∂xi
∣∣∣∣p
(xj) =∂
∂xi(xj ϕ−1)(ϕ(p)) = δij =
1 if i = j,
0 if i 6= j.
Thus
∂∂x1
∣∣p, . . . , ∂
∂xn
∣∣p
are linearly independent elements of TpM . We claim that
they also span TpM and hence form a basis for TpM .
Theorem. Suppose that Mn is a smooth manifold, p ∈ M , (U,ϕ) a smooth coordi-nate system with p ∈ U , and ϕ = (x1, . . . , xn). If v ∈ TpM , then
v =n∑i=1
v(xi)∂
∂xi
∣∣∣∣p
.
Lemma 1 If v ∈ TpM and c ∈ F (p) is constant, then v(c) = 0.
Proof. v(1) = v(1·1) = 1·v(1)+1·v(1) = 2v(1), so v(1) = 0. v(c) = c·v(1) = c·0 = 0.
Lemma 2 Suppose V is a convex neighborhood of 0 in Rn and f : V → R is a smoothfunction. Then there exist smooth functions g1, . . . , gn : V → R such that
1. f = f(0) +∑n
i=1 xigi,
2. gi(0) = ∂f∂xi (0).
Here (x1, . . . , xn) are the standard coordinate functions on Rn.
23
Proof.
f(x1, . . . , xn)− f(0, . . . , 0) =
∫ 1
0
∂
∂t[f(tx1, . . . , txn)] dt
=
∫ 1
0
n∑i=1
∂f
∂xi(tx1, . . . , txn)xi dt
=n∑i=1
xi∫ 1
0
∂f
∂xi(tx1, . . . , txn) dt.
Thus we need only set gi(x1, . . . , xn) =
∫ 1
0∂f∂xi (tx
1, . . . , txn) dt.
Lemma 3 Let (U,ϕ) be a smooth chart on M such that p ∈ U , ϕ(p) = 0, andV = ϕ(U) a convex open neighborhood of 0. If f : U → R is a smooth function,there exist smooth functions g1, . . . , gn : U → R such that
1. f = f(p) +∑n
i=1 xigi,
2. gi(p) = ∂∂xi
∣∣p(f).
Here (x1, . . . , xn) are the components of ϕ.
Proof. Let f = f ϕ−1 and apply Lemma 2.
Proof of Theorem. Assume first that ϕ(p) = 0. Then
v(f) = v(f(p)) +n∑i=1
v(xigi)
= 0 +n∑i=1
v(xi)gi(0)
=n∑i=1
v(xi)∂
∂xi
∣∣∣∣p
(f).
24
If ϕ(p) = (a1, . . . , an), we let yi = xi − ai. Then yi(p) = 0 and ∂∂xi
∣∣p
= ∂∂yi
∣∣∣p. Hence
v(f) =n∑i=1
v(yi)∂
∂yi
∣∣∣∣p
=n∑i=1
v(xi − ai)∂
∂xi
∣∣∣∣p
=n∑i=1
v(xi)∂
∂xi
∣∣∣∣p
(f),
as desired.
Notation. ∂∂xi
∣∣p(f) = ∂f
∂xi (p).
Corollary. If ϕ = (x1, . . . , xn) and ψ = (y1, . . . , yn) are two smooth coordinatesystems at p, then
(a)
∂
∂yi
∣∣∣∣p
=n∑i=1
∂xi
∂yj(p)
∂
∂xi
∣∣∣∣p
(transformation of basis elements),
(b)
v(xi) =n∑j=1
∂xi
∂yj(p)v(yj)
(transformation of components).
Proof.
(a) Apply the Theorem with v = ∂∂yj .
25
(b)
n∑i=1
v(xi)∂
∂xi
∣∣∣∣p
= v
=n∑j=1
v(yj)∂
∂yj
∣∣∣∣p
=n∑
i,j=1
v(yj)∂xi
∂yj(p)
∂
∂xi
∣∣∣∣p
.
Now compare coefficients of ∂∂xi
∣∣p.
Definition. A cotangent vector to Mn at p is a linear functional α : TpM → R.
Let T ∗pM be the space of cotangent vectors at p. T ∗
pM is just the dual vector spaceto TpM .
If f : U → R is a smooth function, U ⊆ M , we define dfp ∈ T ∗pM by dfp(v) = v(f).
dfp is called the differential of f at p. If ϕ = (x1, . . . , xn) is a smooth coordinatesystem defined on some open neighborhood of p, then
dxi∣∣p
(∂
∂xj
∣∣∣∣p
)=
∂
∂xj
∣∣∣∣p
(xi) = δij =
1 if i = j,
0 if i 6= j.
Thus dx1 |p, . . . , dxn |p is the basis for T ∗pM which is dual to
∂∂x1
∣∣p, . . . , ∂
∂xn
∣∣p
.
Proposition. dfp =∑n
i=1∂f∂xi (p)dx
i |p.
26
Proof. If v ∈ TpM ,
df |p (v) = v(f)
=n∑i=1
v(xi)∂
∂xi
∣∣∣∣p
(f)
=n∑i=1
∂f
∂xi(p) dxi |p (v),
as desired.
If α ∈ T ∗pM , the numbers α
(∂∂x1
∣∣p
), . . . , α
(∂∂xn
∣∣p
)are called the components of α
with respect to the coordinates (x1, . . . , xn).
Corollary. If ϕ = (x1, . . . , xn) and ψ = (y1, . . . , yn) are two smooth coordinatesystems at p, then
(a) dyi |p=∑n
j=1∂yi
∂xj (p) dxj |p (transformation of basis elements),
(b) α
(∂∂yj
∣∣∣p
)=∑n
i=1∂xi
∂yj (p)α(
∂∂xi
∣∣p
)(transformation of components).
27
Chapter 6
October 11, 2005
Suppose that F : M → N is a smooth map, p ∈M . Define a linear map F∗p : TpM →TF (p)N by
F∗p(v)(f) = v(f F ),
whenever f : U → R is a smooth function and U an open neighborhood of F (p) inN .
We need to check that F∗p(v) really is a tangent vector and F∗p is linear. Thus weneed to verify that F∗p(v) satisfies the three axioms
1. F∗p(v) is local,
2. F∗p(v) is linear over R,
3. F∗p(v) satisfies the Leibniz rule.
It is straightforward to check each of these. For example, to check (3), we note that
F∗p(v)(fg) = v((fg) F )
= v((f F ) · (g F ))
= (f F )(p)v(g F ) + g F (p)v(f F )
= f(F (p))F∗p(v)(g) + g(F (p))F∗p(v)(g).
Similarly, it is straightforward to check that
F∗p : TpM → TF (p)N
28
is linear.
If ϕ = (x1, . . . , xn) are local coordinates on M about p, and ψ = (y1, . . . , yn) are localcoordinates on N about F (p), then
F∗p
(∂
∂xi
∣∣∣∣p
)=
m∑j=1
(F∗p
(∂
∂xi
∣∣∣∣p
)(yj)
)∂
∂yj
∣∣∣∣F (p)
=m∑j=1
∂
∂xi
∣∣∣∣p
(yj F )∂
∂yj
∣∣∣∣F (p)
=m∑j=1
∂(yj F )
∂xi(p)
∂
∂yj
∣∣∣∣F (p)
.
If
v =(
∂∂x1
∣∣p· · · ∂
∂xn
∣∣p
)a1
...an
,
then
F∗p(v) =
(∂∂y1
∣∣∣F (p)
· · · ∂∂ym
∣∣∣F (p)
)∂(y1F )∂x1 (p) · · · ∂(y1F )
∂xn (p)...
...∂(ymF )∂x1 (p) · · · ∂(ymF )
∂xn (p)
a
1
...an
.
Thus the matrix representing the linear map F∗p : TpM → TF (p)N is the matrixrepresenting the linear map D(ψ F ϕ−1)(ϕ(p)). From this it follows that
(a) F is an immersion if F∗p is injective for all p ∈M ,
1. F is a submersion if F∗p is surjective for all p ∈M .
We can define F ∗p : T ∗
F (p)N → T ∗pM by
F ∗p (α)(v) = α(F∗p(v)).
Then if
α =(a1 · · · am
)dy1 |F (p)...
dym |F (p)
,
29
then
F ∗p (α) =
(a1 · · · am
)∂(y1F )∂x1 (p) · · · ∂(y1F )
∂xn (p)...
...∂(ymF )∂x1 (p) · · · ∂(ymF )
∂xn (p)
dx
1 |p...
dxn |p
.
The components of α ∈ T ∗pM transform like the coordinate basis for TpM . Moreover,
(a) F is an immersion if F ∗p is surjective for all p ∈M ,
(b) F is a submersion if F ∗p is injective for all p ∈M .
Proposition 1 If F : M → N and G : N → P are smooth maps, then the followingdiagrams commute:
TpM(GF )∗p //
F∗p $$IIIIIIIII
TG(F (p))P
TF (p)NG∗F (p)
88rrrrrrrrrr
T ∗pM T ∗
G(F (p))P(GF )∗poo
G∗F (p)yyssssssssss
T ∗F (p)N
F∗p
ccHHHHHHHHH
Proof. (G F )∗p(v)(f) = v(f G F ), while
(G∗F (p) F∗p(v))(f) = F∗p(v)(f G) = v(f G F ).
The two expressions agree, so G∗F (p) F∗p = (G F )∗p. Dualizing yields (G F )∗p =F ∗p G∗
F (p).
Application. Suppose γ : (a, b) →M and γ(t0) = p. Then
γ′(t0) = γ∗t0
(d
dt
∣∣∣∣t0
)∈ TpM
is called the velocity vector of γ at t0. By Proposition 1,
F∗p(γ′(t0)) = F∗p γ∗t0
(d
dt
∣∣∣∣t0
)
= (F γ)∗t0
(d
dt
∣∣∣∣t0
)= (F γ)′(t0).
30
Thus F∗p takes the velocity vector of γ to the velocity vector of F γ. Similarly, F ∗p
takes the differential of f to the differential of f F .
Proposition 2 If F : M → N is a smooth map and f : M → N is a smooth mapand f : U → R is a smooth function defined on an open neighborhood U of p, thenF ∗p (df |F (p)) = d(f F ) |p.
Proof.
F ∗p (df |F (p))(v) = df |F (p) (F∗p(v))
= F∗p(v)(f)
= v(f F )
= d(f F ) |p (v),
as desired.
Recall that we showed that I was a regular value for
F : GL(n,R) → Sym(n,R), F (A) = ATA
by calculating the derivative
F (A+ tB) = (A+ tB)T (A+ tB)
= ATA+ t(BTA+ ATB) + t2BTB
⇒ DF (A)(B)
= BTA+ ATB
and then showing that
DF (A) : M(n,R) → Sym(n,R)
is surjective when F (A) = I.
There is another approach, however, which although slightly more complicated, cansometimes be generalized more easily to other situations.
31
Define xij : GL(n,R) → R and yij : Sym(n,R) → R by
xij
a11 · · · a1
n...
...an1 · · · ann
= aij, yij
b11 · · · b1n...
...bn1 · · · vnn
= bij.
Thus yij = yji , and yij F =∑n
k=1 xki x
kj . Hence
d(yij F ) |I =n∑k=1
xki (I) dxkj |I +
n∑k=1
xkj (I) dxki |I
= dxij |I +dxji |I .
From this we see that F ∗I : TI Sym(n,R) → TI GL(n,R) is injective. If B ∈ F−1(I),
then F LB = F and hence by Proposition 1, F∗B (LB)∗I = F∗I . Since LB is adiffeomorphism, (LB)∗I is an isomorphism and if F ∗
I is injective, then F ∗B is injective
for B ∈ F−1(I). Hence FB∗ is surjective for B ∈ F−1(I), and I is a regular value.
32
Chapter 7
October 13, 2005
Let M be a smooth manifold and f : M → R a real-valued smooth function. Thesupport of f is
supp(f) = closure of p ∈M : f(p) 6= 0.
Definition. Let U = Uα : α ∈ A be an open cover of a smooth manifold M . Asmooth partition of unity subordinate to U is a collection ψα : α ∈ A such that
1. ψα : M → [0, 1] is a smooth function,
2. supp(ψα) ⊆ Uα,
3. If p ∈ M , there is an open neighborhood V of p such that supp(ψα) ∩ V = ∅for all but finitely many α,
4.∑
α∈A ψα = 1.
Theorem. Let U = Uα : α ∈ A be an open cover of a compact smooth manifoldM . Then there is a smooth partition of unity subordinate to U .
Lemma 1 Let p ∈ M and V be an open neighborhood of p. Then there exists asmooth function f : M → [0, 1] such that
(a) f ≡ 1 on a neighborhood of p,
33
(b) supp(f) ⊆ V .
Proof.
1. Define g1 : R → R by
f1(s) =
e−
1s , for s > 0,
0, for s ≤ 0.
We claim that g1 is C∞. Indeed, the only possible bad point is s = 0. But
d
ds
(e−
1s
)= p1
(1
s
)e−
1s ,
where p1 is a polynomial, and by induction,
dk
dsk
(e−
1s
)=
d
ds
(pk−1
(1
s
)e−
1s
)= pk
(1
s
)e−
1s ,
where pk is a polynomial. Thus
lims→0
dk
dsk
(e−
1s
)= lim
s→0pk
(1
s
)e−
1s = lim
t→∞pk(t)e
−t = 0,
lims→0
1
s
dk
dsk
(e−
1s
)= lim
s→0
1
spk
(1
s
)e−
1s = lim
t→∞tpk(t)e
−t = 0.
It follows that g1 is C∞ and dkgdsk (0) = 0 for all k.
2. Let g2(s) = g1(s)g1(1− s).
3. Let
g3(s) =
∫ s0g2(u) du∫ 1
0g2(u) du
.
4. Let g4(s) = g3(2− |s|). Then
g4(s) =
0 for s ≥ 2,
1 for −1 ≤ s ≤ 1,
0 for s ≤ −2.
34
5. For ε > 0, define hε : Rn → R by hε(x) = g4
(|x|ε
). Then
gε(x) =
0 for |x| ≥ 2ε,
1 for |x| ≤ ε,
and supp(gε) ⊆ x ∈ Rn : |x| ≤ 2ε.
6. Let (U,ϕ) be a coordinate system on M so that p ∈ U ⊆ V and ϕ(p) = 0.Choose ε > 0 so that
x ∈ Rn : |x| ≤ 2ε ⊆ ϕ(U),
and define f : M → [0, 1] by
f(q) =
hε(ϕ(q)) for q ∈ U ,
0 for q 6∈ U .
Then f ≡ 1 on a neighborhood of p, and supp(f) ⊆ V .
Lemma 2 Let K be a compact subset of a smooth manifold M and V an open subsetof M containing p. Then there is a smooth function f : M → [0, 1] such that
(a) f > 0 on K,
(b) supp(f) ⊆ V .
Proof. By Lemma 1, if p ∈ K there is a smooth function fp : M → [0, 1] suchthat fp ≡ 1 on a neighborhood of p and supp(fp) ⊆ V . Let Up = q ∈ M :fp(q) > 0. Since K is compact, there are finitely many points p1, . . . , pk such thatV ⊆ Up1 ∪ · · · ∪ Upk
. Let f = 1k(fp1 + · · · + fpk
). Then f has the desired properties.
Lemma 3 Let Uα : α ∈ A be a finite open cover of a smooth manifold M . Thenthere is an open cover Vα : α ∈ A of M such that Vα ⊆ Uα.
35
Proof. For each p ∈ M , choose an open neighborhood Up of p such that Up ⊆ Uαfor some α. By compactness, there exist finitely many points p1, . . . , pk such thatM ⊆ Up1 ∪ · · · ∪ Upk
. Let Vα =⋃Upi
: Upi⊆ Uα. Vα is open, Vα ⊆ Uα, and
Vα : α ∈ A is an open cover of M .
Proof of Theorem. By Lemma 2, there exists a smooth function fα : M → [0, 1]such that fα > 0 on Vα and supp(fα) ⊆ Uα. Let
f =∑α∈A
fα, ψα =fαf.
Then ψα : α ∈ A is a partition of unity subordinate to Uα : α ∈ A.
Proposition. If M is compact and F : M → N is a one-to-one immersion, it is animbedding.
Proof. We need only show that F maps M homeomorphically onto F (M), and in-deed, it suffices to show that if C is closed in M , then F (c) is closed in F (M), whereF (M) is given the subspace topology. But if C is closed in M , then C is closed (aclosed subset of a compact space is compact). Hence F (C) is compact (the continu-ous image of a compact space is compact). Hence F (C) is closed (a compact subsetof a Hausdorff space is closed) if F maps M homeomorphically to F (M) and is animbedding.
Theorem. A compact n-dimensional smooth manifold Mn possesses an imbeddinginto some Euclidean space RN .
Proof. Cover M by finitely many smooth charts
(U1, (x11, . . . , x
n1 )), (U2, (x
12, . . . , x
1n)), . . . , (Um, (x
1m, . . . , x
nm)).
Let ψα : α ∈ A be a partition of unity subordinate to U = Uα : 1 ≤ α ≤ m.Define
F : Mn → (Rn+1)× (Rn+1)× · · · × (Rn+1)︸ ︷︷ ︸m
∼= R(n+1)m
36
by
F = ((ψ1, ψ1x11, . . . , ψ1x
n1 ), (ψ2, ψ2x
12, . . . , ψ2x
n2 ), . . . , (ψm, ψmx
1m, . . . , ψmx
nm)).
F is clearly smooth. We need to check that it is a one-to-one immersion.
F is one-to-one: If F (p) = F (q), then ψj(p) = ψj(q) for 1 ≤ j ≤ m. At least one ofthe ψα(p)’s is nonzero, say ψ1(p) 6= 0. Then ψ1(q) 6= 0, and p and q both lie in U1.Hence (ψ1x
i1)(p) = ψ1x
i1(q) for 1 ≤ i ≤ n, so xi1(p) = xi1(q) for 1 ≤ i ≤ n, and so
p = q, and F is indeed one-to-one.
37
Chapter 8
October 18, 2005
Theorem. If Mn is an n-dimensional smooth manifold, there is an imbeddingF : Mn → RN into some Euclidean space RN .
Recall the proof we started last time: Cover M by finitely many charts
(U1, (x11, . . . , x
n1 )), . . . , (Um, (x
1m, . . . , x
nm)),
let ψ1, . . . , ψm be a smooth partition of unity subordinate to U1, . . . , Um, anddefine F : M → Rn+1 × · · · × Rn+1︸ ︷︷ ︸
m
by
F = ((ψ1, ψ1x11, . . . , ψ1x
n1 ), . . . , (ψm, ψmx
1m, . . . , ψmx
nm)).
Last time we showed that F is one-to-one.
We now need to show that F is an immersion. Let
πj : Rn+1 × · · · × Rn+1︸ ︷︷ ︸m
→ Rn+1
be the projection on the jth factor. It will suffice to show that the rank of the mapping
πj F = (ψj, ψjx1j , . . . , ψjx
nj )
is n at any point of the interior of supp(ψj). We can write
πj F ϕ−1j = (ψ, ψx1, . . . , ψxn),
38
where ψ = ψj ϕ−1j and xi = xij ϕ−1
j . Then
rank(D(πj F ϕ−1j )(ϕj(p))) =
∂ψ∂x1 · · · ∂ψ
∂xn
ψ + x1 ∂ψ∂x1 · · · x1 ∂ψ
∂xn
......
xn ∂ψ∂x1 · · · ψ + xn ∂ψ
∂xn
.
Rank is preserved by elementary row operations. Thus if we subtract xj times thefirst row from the (j + 1)st row for 1 ≤ j ≤ n, we do not change the rank. Hence
rank(D(πj F ϕ−1j )(ϕj(p))) = rank
∂ψ∂x1 · · · ∂ψ
∂xn
ψ · · · 0...
...0 · · · ψ
= n
so long as ψ 6= 0. Thus F is indeed an immersion. Since a one-to-one immersionF : M → RN is an imbedding when M is compact, F is an imbedding.
We would like to show that we can take N = 2n+ 1. To do this we need to make the“tangent bundle” into a smooth manifold.
If M is an n-dimensional smooth manifold, let TM =⊔TpM : p ∈ M (abstract
disjoint union). Define π : TM → M by π(TpM) = p. Let (U,ϕ) be a smoothcoordinate system on M with ϕ = (x1, . . . , xn) and define
x1, . . . , xn : π−1(U) →M
by
xi
(n∑j=1
aj∂
∂xj
∣∣∣∣p
)= ai.
Define ϕ : π−1(U) → R2n by ϕ = (x1 π, . . . , xn π, x1, . . . , xn). If (V, ψ) is a secondsmooth chart on M with ψ = (y1, . . . , yn), we can construct y1, . . . , yn : π−1(Y ) → Rby
yi
(n∑j=1
aj∂
∂yj
∣∣∣∣p
)= ai,
39
and construct a map ϕ : π−1(V ) → R2n by ψ = (y1 π, . . . , yn π, y1, . . . , yn). Ifp ∈ U ∩ V and
v =n∑j=1
aj∂
∂xj
∣∣∣∣p
,
then
yi(v) = yi
(n∑
j,k=1
aj∂yk
∂xj(p)
∂
∂yk
∣∣∣∣p
)
=n∑j=1
aj∂yi
∂xj(p)
=n∑j=1
∂yi
∂xj(p)xj(v),
so y1
...yn
=
∂y1
∂x1 (p) · · · ∂y1
∂xn (p)...
...∂yn
∂x1 (p) · · · ∂yn
∂xn (p)
x
1
...xn
.
Thus we see that the coordinates (y1 π, . . . , yn π, y1, . . . , yn) depend smoothly on(x1 π, . . . , xn π, x1, . . . , xn).
We now give TM the unique topology such that π−1(U) is open for each chart (U,ϕ)and ϕ : π−1(U) → ϕ(U) ⊆ R2n is a homeomorphism. We check that this topology issecond countable and Hausdorff. Hence TM is a smooth manifold of dimension 2n.Moreover, π : TM →M is a smooth map.
Definition. Let Mn be a smooth manifold and U an open subset of M . A smoothvector field on U is a smooth map X : U → TM such that π X = idU .
For example, if (U,ϕ) is a smooth coordinate system on M , ϕ = (x1, . . . , xn), andf 1, . . . , fn : U → R are smooth functions, we can define a vector field
∑ni=1 f
i ∂∂xi on
U by (n∑i=1
f i∂
∂xi
)(p) =
n∑i=1
f i(p)∂
∂xi
∣∣∣∣p
.
Indeed any vector field on U is of this form.
40
Chapter 9
October 20, 2005
A subset A ⊆ Rn is said to have measure zero if for any ε > 0, there is a countablecollection Ci of cubes such that A ⊆
⋃Ci and
∑∞i=1 vol(Ci) < ε.
Definition. A subset A of a smooth manifold M has measure zero if whenever(U,ϕ) is a smooth chart on M , ϕ(U ∩ A) has measure zero in Rn.
We can say that a property holds for almost all p ∈ M if it holds for p ∈ M − A,where A has measure zero.
There is another meaning to the phrase “almost all” that is sometimes used. It isbased on
Baire Category Theorem. Let M be a complete metric space. Then a countableintersection of open dense subsets of M is dense.
Definition. A subset A of a complete metric space M is said to be residual if it isa countable intersection of open dense sets.
Sard’s Theorem. The set of regular values of a smooth map F : M → N is
(a) a set whose complement has measure zero.
41
(b) a residual set.
We will not prove this fundamental theorem. Both parts are proven in Chapter 3 ofHirsch, Differential Topology, Springer, 1976.
Part (b) of the theorem was extended to infinite-dimensional manifolds in 1966.
Note that if dimM < dimN , F∗p : TpM → TF (p)N can never be surjective, so Sard’sTheorem says that almost all points q ∈ N are not in the image of F , or the set ofpoints q not in the image of F is residual.
Compact Whitney Theorem. (1936) Any compact smooth n-dimensional mani-fold Mn possesses an imbedding F : Mn → R2n+1.
Proof. Since we already have that Mn has an imbedding into RN for some N , itsuffices to show that if F : Mn → RN is an imbedding and N > 2n + 1, then thereis an imbedding F0 : Mn → RN−1. Let u be a unit-length vector in RN , and letπu : RN → RN−1, the orthogonal complement to u, be the orthogonal projection. Weclaim that for almost all u ∈ SN−1, or for a residual set of u ∈ N ,
πu F : M → RN−1
is an imbedding. We need only show that
(i) πu is one-to-one,
(ii) πu is an immersion.
πu is one-to-one: If πu F is not one-to-one, there exist points p, q ∈M , p 6= q, suchthat
F (p)− F (q)
|F (p)− F (q)|= u.
Thus u is in the image of the map
G : M ×M −∆ → SN−1, G(p, q) =F (p)− F (q)
|F (p)− F (q)|,
42
where ∆ = (p, q) ∈ M ×M : p = q. If N > 2n + 1, then N − 1 > 2n, so for ubelonging to a residual set, or for almost all u, u is not in the image of G, and henceπu F is one-to-one.
πu F is an immersion: Note that F : Mn → RN induces a map F∗ : TM → RN
defined byF∗(v) = Fp∗(v) ∈ TF (p)RN = RN , for v ∈ TpM.
Moreover, this map is smooth. If πu F is not an immersion, there exists a nonzerovector v ∈ TM such that
πu F∗(v) = 0 or F∗v is parallel to u.
In other words, if πu F is not an immersion, u lies in the image of
H : TM − 0 vectors → SN−1, H(v) =F∗(v)
|F∗(v)|.
But since 2n+ 1 < N ,
dim(TM − 0 vectors) = 2n < N − 1 = dimSN−1.
Hence for u belonging to a residual set, or for almost all u, πu F is an immersion.
It follows from (i) and (ii) that for u belonging to a residual set, or for almost all u,πu F is an imbedding.
Whitney Immersion Theorem. (1936) If Mn is a compact n-dimensional mani-fold, there is an immersion F : Mn → R2n.
Sketch of Proof. We modify part (ii) of the above proof. Since M ⊆ RN , RpM ⊆RN for each p ∈ M . Thus if v ∈ TpM ⊆ TM , we can determine its length |v|. Welet T 1M = v ∈ TM : |v| = 1. One can show that this is a submanifold of TM andthat dimT 1M = 2n − 1. If F : Mn → R2n+1, we can define F ∗ : T 1M → R2n+1 asbefore, as well as
H : T 1M → S2n, H(v) =F∗(v)
|F∗(v)|.
But thendimT 1M = 2n− 1 < 2n = dimS2n,
43
so by Sard’s theorem, for u in a residual set or for almost all u, πuF is an immersion.
9.1 Vector Fields
Let M be a smooth manifold, TM its tangent bundle, and π : TM → M the pro-jection. A smooth vector field X on M is a smooth map X : M → TM such thatπ X = idM .
Recall that in terms of local coordinates (U, (x1, . . . , xn)), vector fields on U are ofthe form
X =n∑i=1
f i∂
∂xi,
where (f 1, . . . , fn) are real-valued functions.
Let X(M) be the set of smooth vector fields on M and F (M) the set of smoothreal-valued functions on M . If X, Y ∈ X(M) and f ∈ F (M), define
X + Y ∈ X(M) by (X + Y )(p) = X(p) + Y (p),
f ·X ∈ X(M) by (fX)(p) = f(p)X(p).
These operations map X(M) into an F (M)-module.
9.1.1 Two Ways of Looking at Vector Fields
Vector Fields are Differential Operators
If f ∈ F (M) and X ∈ X(M), define X(f) ∈ F (M) by
X(f)(p) = X(p)(f).
As a differential operator, X satisfies the following axioms:
1. X(af + bg) = aX(f) + bX(g) for a, b ∈ R, f, g ∈ F (M),
44
2. X(fg) = (Xf)(g) + fX(g) for f, g ∈ F (M).
In terms of local coordinate (x1, . . . , xn) on U ,
X(p) =n∑i=1
X(p)(xi)∂
∂xi
∣∣∣∣p
=n∑i=1
X(xi)(p)∂
∂xi
∣∣∣∣p
,
so
X =n∑i=1
X(xi)∂
∂xi
on U .
Vector Fields are Systems of Ordinary Differential Equations
Definition. An integral curve for a vector field X is a smooth curve γ : (a, b) →Msuch that
γ′(t) = X(γ(t))
for t ∈ (a, b).
Let (U, (x1, . . . , xn)) be a local coordinate system on M , and suppose γ : (a, b) → U ,Then
γ′(t) =n∑i=1
γ′(t)(xi)∂
∂xi
∣∣∣∣γ(t)
=n∑i=1
d(xi γ)dt
(t)∂
∂xi
∣∣∣∣γ(t)
.
If X =∑n
i=1 fi ∂∂xi , then
X(γ(t)) =n∑i=1
f i(γ(t))∂
∂xi
∣∣∣∣γ(t)
.
Define f i : ϕ(u) → R by f i(x1(p), . . . , xn(p)) = f(p). Then
X(γ(t)) =n∑i=1
f i(x1 γ(t), . . . , xn γ(t)) ∂
∂xi
∣∣∣∣γ(t)
.
Hence γ′(t) = X(γ(t)) implies that
d(xi γ)dt
(t) = f i(x1 γ(t), . . . , xn γ(t)), 1 ≤ i ≤ n.
This is just a system of n ordinary differential equations in canonical form.
45
Chapter 10
October 25, 2005
Recall that TM =⊔TpM : p ∈ M is a smooth manifold. A vector field on M is a
smooth map X : M → TM such that π X = idM .
In local coordinates (U,ϕ), where ϕ = (x1, . . . , xn), we write
X =n∑i=1
f i∂
∂xi, where f i : U → R,
each f i being smooth.
An integral curve for X is a smooth curve γ : (a, b) →M such that γ′(t) = X(γ(t))for all t ∈ (a, b).
Initial Value Problem. Given a vector field X on M , p ∈ M , t0 ∈ R, find anintegral curve γ : (a, b) →M for X such that t0 ∈ (a, b) and γ(t0) = p.
In local coordinates (U, (x1, . . . , xn)), we can write
γ′(t) =n∑i=1
γ′(t)(xi)∂
∂xi
∣∣∣∣γ(t)
=n∑i=1
d(xi γ)dt
(t)∂
∂xi
∣∣∣∣γ(t)
,
X(γ(t)) =n∑i=1
f i(x1 γ(t), . . . , xn γ(t)) ∂
∂xi
∣∣∣∣γ(t)
,
46
where f i = f i ϕ−1, so the initial value problem can be written in local coordinatesas
d(x1 γ)dt
(t) = f 1(x1 γ(t), . . . , xn γ(t)),...
d(xn γ)dt
(t) = fn(x1 γ(t), . . . , xn γ(t)),
(x1 γ)(t0) = x1(p0),
...
(xn γ)(t0) = xn(p0),
and this is just the standard initial-value problem from the theory of ordinary differ-ential equations.
Fundamental Existence and Uniqueness Theorem. If X is a smooth vectorfield on M and p0 ∈ M , there is an open neighborhood U of p0 and an ε > 0 suchthat if p ∈ U , the initial value problem
γ′(t) = X(γ(t)), γ(t0) = p
has a unique solution γ : (t0 − ε, t0 + ε) → M . Moreover, the solution dependssmoothly on the initial conditions.
We will explain what the last sentence means shortly.
We will not give a proof of this theorem. For a proof, consult Boothby, page 127 andthe references cited there.
One reference is worth noting: Lang’s book, Differential and Riemannian Manifolds,contains a complete proof for vector fields on infinite-dimensional Banach manifolds.
Definition. A vector field X on M is said to be complete if the initial value problem
γ′(t) = X(γ(t)), γ(0) = p
47
has a unique solution γp : R →M defined for all t.
Next time we will see that any vector field on a compact manifold is complete.
If X is a complete vector field we can define
ϕt : M →M by ϕt(p) = γp(t)
andθ : R×M →M by θ(t, p) = ϕt(p).
Smooth dependence on initial conditions (the last sentence in the theorem) meansthat θ is smooth.
For a general vector field X, there is an open subset W ⊆ R×M with 0×M ⊆ Wsuch that θ is defined on W . Thus in general ϕt(p) = γp(t) is defined only for some tand p.
Lemma. ϕt ϕs = ϕt+s when both sides are defined.
Proof. t 7→ ϕt(ϕs(p)) and t 7→ ϕt+s(p) are integral curves for X which satisfy theinitial conditions s 7→ ϕs(p). By uniqueness they coincide.
Thus if X is complete, t 7→ ϕt is a homomorphism from the additive group of realnumbers into Diff(M), the space of diffeomorphisms of M , a group under compo-sition. ϕt : t ∈ R is the one-parameter group of diffeomorphisms on Mcorresponding to X.
For a general vector field, we only get a local one-parameter group of local diffeomor-phisms ϕt : t ∈ R.
48
The vector field X can be recovered from the one-parameter group:
X(p)(f) = γ′p(0)(f) =d
dt(f γp)(0) =
d
dt(f(ϕt(p)))
∣∣∣∣t=0
.
49
Chapter 11
October 27, 2005
Let M = R2 with coordinates (x, y), and suppose that
X = −y ∂∂x
+ x∂
∂y.
γ : (a, b) →M is an integral curve for X iff(x γ)′(t) = −(y γ)(t),(y γ)′(t) = (x γ)(t).
In this case the one-parameter group of diffeomorphisms is ϕt : t ∈ R, where
ϕt(x, y) = (x cos t− y sin t, x sin t+ y cos t).
ϕt is a counterclockwise rotation through t radians.
For a general vector field X on M , there is a maximal open subset W ⊆ R×M with0 ×M ⊆ W such that the solution γp to the initial value problem
γ′p(t) = X(γp(t)), γp(0) = p,
is defined for (t, p) ∈ W . Moreover, θ : W →M defined by θ(t, p) = γp(t) is smooth.We let ϕt(p) = γp(t), when defined.
We say that X is complete if W = R×M .
50
Theorem. If M is a compact manifold, any vector field X on M is complete.
Proof. Cover M by finitely many open sets U1, . . . , Um such that there are con-stants ε1 > 0, . . . , εm > 0 such that the initial value problem
γ′p(t) = X(γp(t)), γp(0) = p
has a solution γp : (−εi, εi) → M when p ∈ Ui. Let ε = minε1, . . . , εm. Thenϕt : M →M is defined for t ∈ (−ε, ε).
For t ∈ (−Nε,Nε), setϕt = ϕt/N ϕt/N · · · ϕt/N︸ ︷︷ ︸
N
.
If s is sufficiently small, it follows from the lemma ϕt+s = ϕt ϕs that t 7→ ϕt is anintegral curve for X for t ∈ (−Nε,Nε). By uniqueness of integral curves ϕt = ϕt.We can let N →∞ and thereby define ϕt for all t ∈ R.
If X, Y ∈ X(Mn) = smooth vector fields on M we define the Lie bracket [X, Y ] ∈X(Mn) by
[X, Y ](p)(f) = X(Y (f))(p)− Y (X(f))(p).
We need to check that [X, Y ](p) is actually a tangent vector and [X,Y ](p) dependssmoothly on p. This is easiest to do in local coordinates. Suppose
X =n∑i=1
gi∂
∂xi, Y =
n∑i=1
hi∂
∂xi.
Then
[X,Y ](f) =n∑i=1
n∑j=1
[gi
∂
∂xi
(hj∂f
∂xj
)− hi
∂
∂xi
(gj
∂
∂xj
)]
=n∑i=1
n∑j=1
(gi∂hj
∂xi− hi
∂gj
∂xi
)∂
∂xj+
n∑i=1
n∑j=1
gihj(∂
∂xi
(∂f
∂xj
)− ∂
∂xj
(∂f
∂xj
))
=n∑i=1
n∑j=1
(gi∂hj
∂xi− hi
∂gj
∂xi
)∂
∂xj.
51
The following properties of the Lie bracket are easily verified:
[X, Y ] = −[Y,X],
[aX + bY, Z] = a[X,Z] + b[Y, Z],
[[X, Y ], Z] + [[Y, Z], X] + [[Z,X], Y ] = 0,
for X, Y, Z ∈ X(Mn), a, b ∈ R. (These properties imply that [·, ·] makes X(Mn) intoa Lie algebra.)
The Lie bracket is related to the Lie derivative as follows:
First some definitions. If F : M → N is a smooth map, a vector field X on M is saidto be F -related to a vector field X on N if
F∗p(X(p)) = X(F (p)),
where F∗ : TM → TN is the map defined by F∗(v) = F∗p(v) for v ∈ TpM . Equiva-lently,
F∗p(X(p)) = X(F (p)) ⇐⇒ F∗p(X(p))(f) ⇐ X(F (p))(f)
⇐⇒ X(p)(f F ) = X(F (p))(f) ⇐⇒ X(f F ) = (X F )(f),
the last three conditions holding for all functions f . Thus we can say that X isF -related to the vector field X if
X(f F ) = (X F )(f) for all f : N → R.
IfX is a vector field onM , it is not always true that there is a vector field X onN suchthat X is F -related to X. However, if F : M → N is a diffeomorphism we can definea vector field F∗(X) which is F -related to X by (F∗(X))(q) = F∗F−1(q)(X(F−1(q))).
52
Recall that we can differentiate a function f in the direction of X by the formula
X(f) =d
dt(ϕ∗t (f))
∣∣∣∣t=0
.
This is sometimes called the Lie derivative of f in the direction of X.
Definition. If Y is a smooth vector field on M , the Lie derivative of Y in thedirection of X is the vector field LXY defined by
LX(Y )(p) = − d
dt(ϕ∗t (Y )(p))
∣∣∣∣t=0
.
Note that t 7→ ϕ∗t (Y )(p) is a smooth curve in TpM , so we can take its derivative.LXY can be thought of as the directional derivative of Y in the direction of X:
(LXY )(p) =d
dt((ϕ−t)∗(Y )(p))
∣∣∣∣t=0
= limt→0
1
t[(ϕ−t∗(Y ))(p)− Y (p)]
= limt→0
1
t[(ϕ−t)∗(Y (ϕt(p)))− Y (p)].
Theorem. LXY = [X,Y ].
The proof is based upon the following
Lemma. Suppose X is a smooth vector field on M and X(p) 6= 0. Then there is asmooth coordinate system (U, (x1, . . . , xn)) on M with p ∈ U such that
X =∂
∂x1on U.
53
We will prove the lemma next time. Here is the proof of the theorem:
Assume first that X(p) 6= 0. Choose local coordinates as in the lemma and letϕt : t ∈ R be the one-parameter group of X. Then
x1 = ϕt = x1 + t, xi ϕt = xi for 2 ≤ i ≤ n.
Suppose that Y =∑f i ∂
∂xi . Then
(ϕt)∗(Y )(xi) = Y (xi ϕt) ϕ−1t = Y (xi) ϕ−1
t = f i ϕ−1t ,
so
(ϕt)∗(Y ) =n∑i=1
(f i ϕ−1t )
∂
∂xi,
and
− d
dt(ϕt∗(Y )(p))
∣∣∣∣t=0
= − d
dt
(∑f i ϕ−1
t (p)∂
∂xi
∣∣∣∣p
)
=d
dt
(n∑i=1
f i ϕt∂
∂xi
∣∣∣∣p
),
while
[X, Y ](p) =
[∂
∂x1
n∑i=1
f i∂
∂xi
]
=n∑i=1
∂f i
∂x1
∂
∂xi
∣∣∣∣p
=d
dt
(∑f i ϕt(p)
∂
∂xi
∣∣∣∣p
)∣∣∣∣∣t=0
.
The two expressions are equal.
If X(p) ≡ 0 on a neighborhood of p, then LXY and [X, Y ] are both zero. If X(p) ≡ 0,but there is a sequence pi → p such that X(pi) 6= 0, then LXY (p) = [X, Y ](p) bycontinuity.
54
Chapter 12
November 1, 2005
Recall that if F : M → N is a smooth map, a vector field X on M is F -related to avector field X on N if
F∗p(X(p)) = X(F (p)) for p ∈M.
Note that
F∗p(X(p)) = X(F (p)) ⇐⇒ F∗p(X(p))(f) = X(F (p))(f) for all f ∈ F (F (p))
⇐⇒ X(p)(f F ) = X(F (p))(f) ⇐⇒ X(f F ) = (X F )(f).
So X is F -related to X iff
X(f F ) = (X F )(f) for all f : M → R. (∗)
Finally, X is F -related to X iff F takes integral curves for X to integral curves forX.
Proposition. If X and Y are F -related to X and Y , then [X, Y ] is F -related to
[X, Y ].
The proof is a straightforward exercise using (∗).
55
If F : M → N is a diffeomorphism, and X is a vector field on M , we can define avector F∗(X) on N such that X is F -related to F∗(X) by
(I) F∗(X)(q) = F∗F−1(q)(X(F−1(q))),
(II) (F∗(X)(f)) F = X(f F ) or F∗(X)(f) = X(f F ) F−1,
or
(III) If γ : (a, b) → M is an integral curve for X, F γ : (a, b) → N is an integralcurve for F∗(X).
In particular, if ϕt : t ∈ R is a local one-parameter group corresponding to X, wecan define (ϕt)∗(Y ) by
(ϕt)∗(Y )(q) = (ϕt)∗(Y (ϕ−1t (q))).
Note thatt 7→ (ϕ−t)∗(Y (ϕt(p))) ∈ TpM.
Theorem.
[X, Y ] =d
dt((ϕ−t)∗(Y ))
∣∣∣∣t=0
def= LXY.
To prove this, we need
Lemma. If X is a smooth vector field on M and X(p) 6= 0, there is a smoothcoordinate system (x1, . . . , xn) on a neighborhood U of p such that
X =∂
∂x1.
56
Proof. It is easy to construct coordinates ψ = (y1, . . . , yn) on a neighborhood of p
such that yi(p) = 0 and X(p) = ∂∂y1
∣∣∣p. Let ϕt : t ∈ R be the local one-parameter
group corresponding to X, and let
F (t1, . . . , tn) = ψ ϕt1(ψ−1(0, t2, . . . , tn)).
Then DF (0) = I, so it follows by the inverse function theorem that there exists alocal inverse G. Finally, let η = G ψ, η = (x1, . . . , xn). We claim that (x1, . . . , xn)are the desired coordinates. To see this, define ϕt : Rn → Rn by ϕt(x
1, . . . , xn) =(x1 + t, x2, . . . , xn). Then
η ϕt η−1(x1, . . . , xn) = η ϕt ψ−1 F (x1, . . . , xn)
= η ϕt ϕx1(ψ−1(0, x2, . . . , xn))
= η ϕt+x1(ψ−1(0, x2, . . . , xn))
= η ψ−1 F (x1 + t, x2, . . . , xn)
= (x1 + t, . . . , xn).
So in the η-coordinates X is represented by the one-parameter group ϕt, and thereforeX = ∂
∂x1 .
Proof of Theorem. Suppose X(p) 6= 0. Then in terms of (x1, . . . , xn),
x1 ϕt = x1 + t, x2 ϕt = x2, . . . , xn ϕt = xn.
Suppose now that Y =∑n
i=1 fi ∂∂xi . Then
[X,Y ] =n∑i=1
∂f i
∂x1
∂
∂xi. (∗∗)
On the other hand, since f i = Y (xi),
(ϕt)∗(Y )(xi) = Y (xi ϕt) ϕ−1t = Y (xi) ϕ−1
t = f i ϕ−1t ,
and hence
(ϕt∗)(Y )(p) =∑
f i ϕ−1t (p)
∂
∂xi
∣∣∣∣p
.
57
It follows that
d
dt(ϕt∗(Y ))(p)
∣∣∣∣t=0
=n∑i=1
d
dt(f i ϕ−1
t )(p)∂
∂xi
∣∣∣∣p
= −n∑i=1
d
dt(f i ϕt(p))
∂
∂xi
∣∣∣∣p
= −n∑i=1
∂f i
∂x1(p)
∂
∂xi
∣∣∣∣p
.
Comparison with (∗∗) yields [X, Y ] = − ddt
(ϕt∗(Y ))∣∣t=0
.
If X ≡ 0 in a neighborhood of p, the same formula holds trivially. If X(p) = 0 butthere is a sequence qi → p such that X(qi) 6= 0, the formula holds by continuity.
Corollary. [X, Y ] = 0 iff the one-parameter groups ϕs and ψs of X and Ycommute: ϕt ψs = ψs ϕt.
Proof. ⇐: If ϕtψs = ψsϕt, then (ϕt)∗(Y ) = Y , and hence [X, Y ] = − ddt
(ϕt)∗(Y ) =0.
⇒: Since (ϕt)∗(X) = X, it follows from the Proposition at the beginning of the lec-ture that [X, (ϕt)∗(Y )] = 0, and hence d
dt(ϕt∗(Y )) = 0 for all t, so (ϕt)∗(Y ) = Y . But
then ϕt takes integral curves for Y to integral curves for Y , so ϕt ψs = ψs ϕt.
12.1 The Cotangent Bundle and Differential One-
Forms
Let T ∗M =⊔T ∗
pM : p ∈M (disjoint union) and define π : TM →M by π(TpM) =p. Suppose (Uα, ϕα) : α ∈ A is a smooth atlas on M . Define
pαi : π−1(Uα) → R bypαi
(n∑j=1
aj dxjα
)= ai.
58
We can then give T ∗M the smooth manifold structure defined by the charts
(π−1(Uα), (x1α π, . . . , xnα π, pα1, . . . , pαn)) : α ∈ A.
We can check that the transition functions are smooth by means of the formula forchange of basis of T ∗
pM .
Definition. A smooth one-form on M is a smooth map α : M → T ∗M such thatπ α = idM .
Example. If f : M → R is a smooth map, we can define df : M → T ∗M bydf(p) = df |p. This is called the differential of f .
Suppose F : M → N is a smooth map and α : N → TN is a smooth one-form on N .We can define a smooth one-form F ∗(α) on M by
F ∗(α)p = F ∗p (α(F (p))).
Important Fact. Even though vector fields cannot always be pushed forward undersmooth maps, smooth one-forms can always be pulled back.
12.2 Vector Bundles
Definition. A (smooth) real vector bundle of rank k over M is a triple ξ =(E, π,M), where E and M are smooth manifolds and π : E → M is a smooth map,with the following additional structure:
1. For each p ∈ M , Ep = π−1(p) has the structure of a k-dimensional real vectorspace.
2. (Local triviality) There is an open cover Uα : α ∈ A of M and smooth smooth
59
ψα : π−1(Uα) → Uα × Rk such that the diagram
π−1(Uα)ψα //
π%%LLLLLLLLLL
Uα × Rk
π1xxrrrrrrrrrr
π−1(Uα)
commutes, π1 being the projection on the first factor.
3. If π2 : Uα → Rk → Rk is the projection on the second factor, then ηα = π2 ψαmaps each Ep isomorphically to the vector space Rk.
E is the total space, B the base space, and Ep the fiber.
Examples.
1. The trivial bundle θk = (M × Rk, π1,M).
2. The tangent bundle τ(M) = (TM, π,M).
3. The cotangent bundle τ ∗(M) = (T ∗M,π,M).
In the third case, if (Uα1 , (x1α, . . . , x
nα)) : α ∈ A is a smooth atlas on M , we define
ηα : π−1(Uα) → Rn by ηα = (pα1, . . . , pαn)
andψα : π−1(Uα) → Uα × Rn by ψα = (π, ηα).
Definition. If ξ = (E, π,M) is a smooth vector bundle, a section of ξ is a smoothmap σ : M → E such that π σ = idM .
60
Chapter 13
November 3, 2005
Recall that a real vector bundle of rank k is a triple ξ = (E, π,B), where E and Bare smooth manifolds and π : E → B is a smooth map with extra structure:
1. For each p ∈ B, Ep = π−1(p) has the structure of a k-dimensional real vectorspace.
2. There is an open cover Uα : α ∈ A of M , together with smooth maps ψα :π−1(Uα) → Uα × Rk such that the diagram
π−1(Uα)ψα //
π$$H
HHHHHHHHUα × Rk
π1zzvvvvvvvvv
Uα
commutes.
3. If ψ = (π, ηα), ηα |Ep : Ep → Rk is a vector space isomorphism.
The maps ψα are called vector bundle charts.
If ξ1 = (E1, π1, B1) and ξ2 = (E2, π2, B2) are two smooth vector bundles, a vectorbundle map F : ξ1 → ξ2 is a pair of maps FE : E1 → E2 and FB : B1 → B2 suchthat
61
1. the diagram
E1FE //
π1
E2
π2
B1 FB
// B2
commutes, and
2. FB takes E1p = π−11 (p) linearly to E2F (p) = π−1
2 (F (p)) for each p ∈ B1.
Example. As we saw last time, an important example of a vector bundle is thetangent bundle τ(M) = (TM, π,M) of a smooth manifold M . A smooth map F :M → N induces a vector bundle map F∗ : τ(M) → τ(N) with FB = F and FE = F∗,the map defined by
F∗(v) = F∗p(v) for v ∈ TpM.
In fancy language, we have a “category” of vector bundles and vector bundle maps.
Two vector bundles ξ1 and ξ2 are isomorphic if there are vector bundle mapsF : ξ1 → ξ2 and G : ξ2 → ξ1 such that F G is the identity and G F is theidentity.
Two vector bundles ξ1 = (E1, π1, B) and ξ2 = (E2, π2, B) are isomorphic over B ifthere is a vector bundle isomorphism F : ξ1 → ξ2 with FB the identity on B.
Definition. A map φ : TpM × · · · × TpM︸ ︷︷ ︸k
→ R is k-multilinear if it is linear in each
variable, when the other vectors are fixed.
If k = 2, multilinear is the same as bilinear.
φ(av1 + v2, w) = aφ(v1, w) + φ(v2, w),
φ(v, aw1 + w2) = aφ(v, w1) + φ(v, w2).
62
Let⊗k T ∗M denote the linear space of all k-multilinear maps φ : TpM×· · ·×TpM →
R. If φ ∈⊗k T ∗
pM and ψ ∈⊗` T ∗
pM , we can define φ⊗ ψ ∈⊗k+` T ∗
pM by
(φ⊗ ψ)(v1, . . . , vk+`) = φ(v1, . . . , vk)ψ(vk+1, . . . , vk+`).
This multiplication is bilinear and associative:
(aφ1 + φ2)⊗ ψ = aφ1 ⊗ ψ + φ2 ⊗ ψ,
φ⊗ (aψ1 + ψ2) = aφ⊗ ψ1 + φ⊗ ψ2,
(φ⊗ ψ)⊗ ω = φ⊗ (ψ ⊗ ω).
In particular, we can write φ⊗ ψ ⊗ ω with no danger of confusion.
If (U, (x1, . . . , xn)) is a smooth coordinate system on M , we can define
dxi1 |p ⊗ · · · ⊗ dxik |p∈⊗k
T ∗pM,
and
(dxi1 |p ⊗ · · · ⊗ dxik |p)
(∑j1
aj11∂
∂xj1
∣∣∣∣p
, . . . ,∑jk
ajkk∂
∂xjk
∣∣∣∣p
)= aj11 · · · a
jkk .
Theorem. dxi1 |p ⊗ · · · ⊗ dxik |p: 1 ≤ i1, . . . , ik ≤ n is a basis for⊗k T ∗
pM . Hence⊗k T ∗pM has dimension nk.
Proof. Exercise.
If F : M → N is a smooth map, we can define a linear map
F ∗p :⊗k
T ∗F (p)N →
⊗kT ∗pM
byF ∗p (φ)(v1, . . . , vk) = φ(Fp∗v1, . . . , Fp∗vk).
63
Let⊗k T ∗M =
⊔⊗k T ∗pM : p ∈M
, π :
⊗k T ∗M → M the projection such that
π(T ∗pM) = p.
If (U, (x1, . . . , xn)) is a smooth coordinate system on M , define pi1,...,ik : π−1(U) → Rby
pi1···ik
(n∑
j1=1
· · ·n∑
jk=1
aj1,...,jk dxj1 |p ⊗ · · · ⊗ dxjk |p
)= ai1···ik .
Define η : π−1(U) → Rnkby
η = (p1···1, . . . , pi1···ik , . . . , pn···n),
and define ψ : π−1(U) → U × Rnkby ψ = (π, η). Then ψ : π−1(U) → U × Rnk
is a
vector bundle chart for a rank nk vector bundle⊗k τ ∗M =
(⊗k T ∗M,π,M).
Recall that if ξ = (E, π,M) is a smooth vector bundle, a smooth section of ξ is asmooth function σ : M → E such that π σ is the identity. Let
Γ(ξ) = space of smooth sections of ξ.
If σ ∈ Γ(ξ) and f ∈ F (M), we can define f · σ ∈ Γ(ξ) by (f · σ)(p) = f(p)σ(p). Thisoperation makes Γ(ξ) into an F (M)-module.
A smooth section φ of⊗k τ ∗M is called a covariant tensor field of rank k.
If (U, (x1, . . . , xn)) are local coordinates on M , we can define dxi1 ⊗ · · · ⊗ dxik ∈Γ(⊗k τ ∗M |U
)by
dxi1 ⊗ · · · ⊗ dxik(p) = dxi1 |p ⊗ · · · ⊗ dxik |p .
If φ ∈ Γ(⊗k τ ∗M
)and (U, (x1, . . . , xn)) are smooth coordinates on M ,
φ |U=n∑i=1
· · ·n∑
ik=1
fi1···ik dxi1 ⊗ · · · ⊗ dxik ,
64
where fi1···ik : U → R are smooth.
Suppose F : M → N is a smooth map and φ ∈ Γ(⊗k τ ∗N
). We can then define
F ∗φ ∈ Γ(⊗k τ ∗M
)by
F ∗(φ)(p) = F ∗p (φF (p)).
This defines a linear map F ∗ : Γ(⊗k τ ∗N
)→ Γ
(⊗k τ ∗M).
Definition. A Riemannian metric on M is a smooth section g : M →⊗2 T ∗M
such that
1. g(p)(v, w) = g(p)(w, v) for all v, w ∈ TpM ,
2. g(p)(v, v) ≥ 0, with equality holding only if v = 0.
In terms of local coordinates (U, (x1, . . . , xn)), we can write g |U=∑n
i=1
∑nj=1 gij dx
i⊗dxj. The matrix (gij) satisfies the two conditions:
1. it is symmetric,
2. it is positive definite.
Example. RN has a Euclidean metric
gE = dx1 ⊗ dx1 + · · ·+ dxN ⊗ dxN .
In this case
gIJ = δIJ =
1 if I = J,
0 if I 6= J ,
where I and J range from 1 to N . If v, w ∈ TpRN ,
g(p)(v, w) = v · w,
where the dot is the usual dot product.
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If F : Mn → RN is an immersion, F ∗gE is a Riemannian metric on M , called theinduced metric.
Theorem. Every smooth manifold M possesses a smooth Riemannian metric.
Proof. By Whitney’s imbedding theorem, there is an imbedding F : Mn → RN forsome N . The induced metric g = F ∗(gE) is a Riemannian metric on M .
If (Mn, g) is a Riemannian manifold, an imbedding F : Mn → RN is isometric ifg = F ∗(gE).
In 1956, John Nash proved the following celebrated theorem:
Theorem. Any Riemannian manifold (Mn, g) possesses an isometric imbedding intosome Euclidean space RN .
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Chapter 14
November 8, 2005
Partition of Unity Theorem. (Boothby, pages 186–188) Let Uα : α ∈ A be anopen cover of a smooth manifold M . Then for each α ∈ A, there is a smooth functionψα : M → [0, 1] such that the collection ψα : α ∈ A satisfies:
1. supp(ψα) ⊆ Uα.
2. If p ∈ M , there is an open neighborhood V of p such that supp(ψα) ∩ V = ∅for all but finitely many α.
3.∑
α∈A ψα ≡ 1.
Such a collection ψα : α ∈ A is called a partition of unity subordinate toUα : α ∈ A.
Note. We proved this theorem when M is compact.
Typical Application
Theorem. Every smooth manifold M has a Riemannian metric g.
Proof. Cover M by coordinate systems (Uα, (x1α, . . . , x
nα)) : α ∈ A. Let gα =∑n
i=1 dxiα ⊗ dxiα, a Riemannian metric on Uα. Let ψα : α ∈ A be a partition of
unity subordinate to Uα : α ∈ A. Then g =∑
α∈A ψαgα is a Riemannian metric onM .
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What makes this proof work is that the collection of bilinear maps g : TpM → TpM →R which are symmetric and positive-definite forms a convex set.
Definition. A pseudo-Riemannian metric on M is a section g ∈⊗2 τ ∗M such
that for p ∈M ,
1. g(p) is symmetric,
2. and nondegenerate.
The second condition means that if v ∈ TpM and g(p)(v, ·) is the zero linear func-tional, then v = 0. Equivalently, if g =
∑ni,j=1 gij dx
i ⊗ dxj on U , then det(gij) 6= 0.
A theorem of linear algebra (see Mac Lane and Birkhoff, Algebra, page 387) statesthat if p ∈ M , local coordinates (x1, . . . , xn) can be chosen on a neighborhood of psuch that
(gij(p)) =
(Ik 00 −I`
).
If M is connected, the pair of integers (k, `) does not depend on the choice of p. Inthis case, we say that (k, `) is the signature of the pseudo-Riemannian manifold(M, g).
A pseudo-Riemannian manifold of signature (n, 0) is a Riemannian manifold. Apseudo-Riemannian manifold of signature (n− 1, 1) is a Lorentz manifold.
Not all manifolds have Lorentz metrics. The proof via partitions of unity doesn’twork because symmetric matrices of signature (n− 1, 1) do not form a convex set.
A connected Riemannian manifold can be made into a metric space. Here’s how: Ifγ : [a, b] →M is a smooth curve, we define the length of γ to be
L(γ) =
∫ b
a
√g(γ′(t), γ′(t)) dt.
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If p and q are points of M , we let
d(p, q) = infL(γ) | γ : [a, b] →M is a smooth curve with γ(a) = p, γ(b) = q.
Clearly
1. d(p, q) = d(q, p).
2. d(p, q) + d(q, r) ≤ d(p, r).
3. d(p, q) ≥ 0.
It is a little harder to prove that
4. d(p, q) = 0 ⇒ p = q.
We will do this in a relatively elementary fashion later in the course. For now, wehave the Nash imbedding theorem at our disposal. In RN with the Euclidean metricgE =
∑NI=1 dx
I ⊗ dxI , it is relatively easy to check axiom 4. Indeed we claim thatd(p, q) ≥ |p− q|, where |p− q| is the length of the line from p to q.
We can suppose that p = (0, . . . , 0) and q = (c, 0, . . . , 0). If γ : [a, b] → RN is anycurve from p to q,
L(γ) =
∫ b
a
√gE(γ′(t), γ′(t)) dt
= · · ·
=
∫ b
a
√(dx1
dt
)2
+ · · ·+(dxN
dt
)2
dt
≥∫ b
a
√(dx1
dt(t)
)2
dt
≥∫ b
a
dx1
dtdt
= c− 0
= |p− q|.
So if p 6= q, then infL(γ) : γ is a curve from p to q > 0, establishing axiom 4.
69
If (Mn, g) is any Riemannian manifold, the Nash imbedding theorem states that thereis an imbedding F : Mn → RN such that g = F ∗gE, for some N . If γ : [a, b] → Mis a smooth curve with γ(a) = p and γ(b) = q, then F γ : [a, b] → RN is a smoothcurve with F γ(a) = F (p) and F γ(b) = F (q). Moreover,
L(F γ) =
∫ b
a
√gE((F γ)′(t), (F γ)′(t)) dt
=
∫ b
a
√F ∗gE(γ′(t), γ′(t)) dt
=
∫ b
a
√g(γ′(t), γ′(t)) dt
= L(γ).
Hence L(γ) = L(F γ) ≥ |F (p)− F (q)|. If p 6= q, F (p) 6= F (q) and L(γ) is boundedbelow by a positive constant. So p 6= q implies that d(p, q) > 0, establishing axiom 4.
Thus (M,d) is a metric space. One can show that the metric space topology agreeswith the manifold topology.
Minkowski space-time is Rn+1 with coordinates (t, x1, . . . , xn) and the Lorentzmetric
g = −c2 dt⊗ dt+n∑i=1
dxi ⊗ dxi.
Minkowski space-time is the arena for special relativity. Points of Minkowski space-time are interpreted as events which occur at a given point in space.
A moving object, such as a planet, occupies a curve of events called a world-line.
A vector v ∈ TpM is
timelike if g(v, v) < 0,
lightlike if g(v, v) = 0,
spacelike if g(v, v) > 0.
Suppose γ(t) = (t, x1(t), . . . , xn(t)).
Then γ′(t) is timelike if
−c2 · 1 +n∑i=1
(dxi
dt(t)
)2
< 0;
70
that is, if∑n
i=1
(dxi
dt
)2
< c2.
A timelike curve γ : [a, b] →M is one such that γ′(t) is timelike for all t.
Particles which move slower than light have world lines which are timelike.
A basic axiom of special relativity is that time measured by clocks moving alongthe timelike curve γ : [a, b] → Rn+1 is given by the formula elapsed time along γ is
equal to 1cL(γ), where L(γ) =
∫ ba
√−g(γ′(t), γ′(t)) dt, g being the Lorentz metric.
If γ : [0, τ ] → Rn+1 is defined by γ(t) = (t, 0, . . . , 0), then L(γ) =∫ τ
0c dt = cτ , so the
elapsed time is 1cL(γ) = τ .
The twin paradox is a reflection of the fact that if γ1 and γ2 are two timelike curvesfrom (0, . . . , 0) to (τ, . . . , 0), it is not necessarily the case that 1
cL(γ1) = 1
cL(γ2). For
example, if γ1 is the straight line from (0, . . . , 0) to (τ, . . . , 0), 1cL(γ1) = τ , but if γ2 is
a curve from (0, . . . , 0) to (τ, 0, . . . , 0) such that g(γ′(t), γ′(t)) is closed to zero for allt, then 1
cL(γ2) will also be closed to zero. Thus an astronaut who travels at almost
the speed of light to Alpha Centauri and back (4.3 light years each way) will measurealmost no elapsed time, while his twin who stays on earth will age by about 8.6 years.This may seem paradoxical, but it is the way time behaves in special relativity.
References: Misner, Thorne, and Wheeler, Gravitation. Wald, General Relativity.
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Chapter 15
November 10, 2005
Let ΛkT ∗pM denote the vector space of alternating or skew-symmetric multilinear
mapsφ : TpM × · · · × TpM︸ ︷︷ ︸
k
→ R.
By alternating, we mean that if i 6= j,
φ(v1, . . . , vi, . . . , vj, . . . , vk) = −φ(v1, . . . , vj, . . . , vi, . . . , vk).
Alternate description: Let Sk = bijections from 1, . . . , k to itself, a group undercomposition. Sk is called the symmetric group. Define a group homomorphism
sgn : Sk → ±1 by sgn(σ) =∏i<j
σ(i)− σ(j)
i− j.
A k-multilinear map φ is alternating iff
φ(vσ(1), . . . , vσ(k)) = sgnσφ(v1, . . . , vk) for σ ∈ Sk.
We define a map Alt :⊗k T ∗
pM → ΛkT ∗pM by
Alt(φ)(v1, . . . , vk) =1
k!
∑σ∈Sk
sgnσφ(vσ(1), . . . , vσ(k)).
72
Thus if k = 2, for example,
Alt(φ)(v1, v2) =1
2[φ(v1, v2)− φ(v2, v1)].
It is readily checked that Alt Alt = Alt, so Alt is a projection.
ΛkT ∗pM is a linear subspace of
⊗k T ∗pM . If φ ∈ ΛkT ∗
pM and ψ ∈ Λ`T ∗pM , we define
φ ∧ ψ ∈ Λk+`T ∗pM by φ ∧ ψ =
(k + `)!
k!`!Alt(φ⊗ ψ)
or (alternate definition used by some authors)
φ ·∧ψ ∈ Λk+`T ∗pM by φ ·∧ψ = Alt(φ⊗ ψ).
φ ∧ ψ is called the wedge product of φ and ψ. Both ∧ and ·∧ satisfy the conditions
1. bilinearity:
(aφ+ φ) ∧ ψ = aφ ∧ ψ + φ ∧ ψφ ∧ (aψ + ψ) = aφ+ ∧ψ + φ ∧ ψ.
skew-symmetry: φ ∧ ψ = (−1)k`ψ ∧ φ, when φ ∈ ΛkT ∗pM and ψ ∈ Λ`T ∗
pM .
2. associativity: (φ ∧ ψ) ∧ ω = φ ∧ (ψ ∧ ω).
Only (3) is hard to verify (see Boothby pages 203–205). It is easier to prove associa-tivity for ·∧; one checks that
Alt(Alt(φ⊗ ψ)⊗ ω) = Alt(φ⊗ ψ ⊗ ω) = Alt(φ⊗ Alt(ψ ⊗ ω)).
This shows that (φ ·∧ψ) ·ω = φ ·∧(ψ ·∧ω), so we can write φ ·∧ψ ·∧ω without danger ofconfusion. If
φ ∈ ΛkT ∗pM, ψ ∈ Λ`T ∗
pM, ω ∈ ΛmT ∗pM,
(φ ∧ ψ) ∧ ω =(k + `)!
k!`!(φ ·∧ψ) ·∧ω
=(k + `)!
k!`!
(k + `+m)!
(k + `)!m!φ ·∧ψ ·∧ω
=(k + `+m)!
k!`!m!φ ·∧ψ ·∧ω.
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Similarly,
φ ∧ (ψ ∧ ω) =(k + `+m)!
k!`!m!φ ·∧ψ ·∧ω,
establishing associativity.
Thus we can write φ ∧ ψ ∧ ω with no danger of confusion.
Proposition. If (x1, . . . , xn) is a smooth coordinate system on M , then
dxi1 |p ∧ · · · ∧ dxik |p: 1 ≤ i1 < i2 < · · · < ik ≤ n
is a basis for ΛkT ∗pM . Thus dim ΛkT ∗
pM =(nk
).
Note that ΛkT ∗pM = 0 when k > n.
Let Λ∗TpM = R⊕T ∗pM ⊕Λ2T ∗
pM ⊕· · ·⊕ΛnT ∗pM . The wedge product makes Λ∗TpM
into an associative algebra over R called the exterior algebra.
We let
ΛkT ∗M =⊔ΛkT ∗
pM : p ∈M,
Λ∗T ∗M =⊔Λ∗T ∗
pM : p ∈M,
the total spaces of vector bundles Λkτ ∗M and Λ∗τ ∗M over M .
Definition. A differential k-form on M is a smooth section of Λkτ ∗M ; that is, asmooth map ω : M → ΛkT ∗M such that π ω = id. Let Ak(M) = smooth k-formson M.
If (U, (x1, . . . , xn)) is a smooth coordinate system onM , we can define dxi1∧· · ·∧dxik ∈Ak(U) by
(dxi1 ∧ · · · ∧ dxik)(p) = dxi1 |p ∧ · · · ∧ dxik |p .
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An element ω ∈ Ak(U) can be written as
ω =∑
i1<···<ik
fi1···ik dxi1 ∧ · · · ∧ dxik ,
where fi1···ik : U → R is a smooth function for each choice of i1 < · · · < ik.
If ω ∈ Ak(M) and ψ ∈ A`(M), define φ∧ψ ∈ Ak+`(M) by (ω ∧ψ)(p) = ω(p)∧ψ(p).
Note that A0(M) = F (M) = smooth functions f : M → R.
Theorem. There exists a unique collection of R-linear maps d : Ak(M) → Ak+1(M)such that
1. if f ∈ A0(M), d(f) = df , the differential of f previously defined.
2. d d = 0.
3. if ω ∈ Ak(M) and φ ∈ A`(M),
d(ω ∧ φ) = dω ∧ φ+ (−1)k ω ∧ dφ.
d is called the exterior derivative.
For example, suppose M = R3 with the usual coordinates (x, y, z). If f ∈ A0(M),df = ∂f
∂xdx+ ∂f
∂ydy + ∂f
∂zdz.
If ω ∈ A1(M), say ω = P dx+Q dy +R dz,
dω = dP ∧ dx+ dQ ∧ dy + dR ∧ dz
=
(∂P
∂xdx+
∂P
∂ydy +
∂P
∂zdz
)∧ dx
+
(∂Q
∂xdx+
∂Q
∂ydy +
∂Q
∂zdz
)∧ dy
+
(∂R
∂xdx+
∂R
∂ydy +
∂R
∂zdz
)∧ dz
=
(∂R
∂y− ∂Q
∂z
)dy ∧ dz +
(∂P
∂z− ∂R
∂x
)dz ∧ dx+
(∂Q
∂x− ∂P
∂y
)dx ∧ dy.
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If ω ∈ A2(M), say ω = P dy ∧ dz +Q dz ∧ dx+R dx ∧ dy,
dω = dP ∧ dy ∧ dz + dQ ∧ dz ∧ dx+ dR ∧ dx ∧ dy
=
(∂P
∂xdx+
∂P
∂ydy +
∂P
∂zdz
)∧ dy ∧ dz(
∂Q
∂xdx+
∂Q
∂ydy +
∂Q
∂zdz
)∧ dz ∧ dx(
∂R
∂xdx+
∂R
∂ydy +
∂R
∂zdz
)∧ dx ∧ dy
=
(∂P
∂x+∂Q
∂y+∂R
∂z
)dx ∧ dy ∧ dz.
A0(R3)d−−−−→
gradientA1(R3)
d−−−→curl
A2(R3)d−−−−−−→
divergenceA3(R3).
Lemma. If d exists, it is “local”:
supp(dω) ⊆ supp(ω).
Proof. We show that if p 6∈ supp(ω), p 6∈ supp(dω). Suppose ω ≡ 0 on a neigh-borhood V of p. Let f : Mn → [0, 1] be a smooth function such that f ≡ 1 on aneighborhood W of p (with W ⊆ U) and f ≡ 0 on M − V . Then fω ≡ ω. If q ∈ W ,
0 = d(fω)(q) = df(q) ∧ ω(q) + f(q) dω(q) = dω(q),
so W ∩ supp(dω) = ∅.
Proof of Theorem. Uniqueness: The Lemma implies that it suffices to showuniqueness when M = U , the domain of a smooth coordinate system (x1, . . . , xn). Ifω ∈ Ak(U), then
ω =∑
i1<···<ik
fi1···ik dxi1 ∧ · · · ∧ dxik .
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By the axioms for d, we must have
dω =∑
i1<···<ik
dfi1···ik ∧ dxi1 ∧ · · · ∧ dxik +∑
i1<···<ik
fi1···ik d(dxi11 ∧ · · · ∧ dxik).
Axioms (2), (3), and induction show that d(dxi1 ∧ · · · ∧ dxik) = 0. Thus
d
( ∑i1<···<ik
fi1···ik dxi1 ∧ · · · ∧ dxik
)=
∑i1<···<ik
dfi1···ik ∧ dxi1 ∧ · · · ∧ dxik . (∗)
This establishes uniqueness.
Local Existence: We define d by (∗) and check that it satisfies the axioms. Clearly, dis R-linear and satisfies Axiom 1. Axiom 3: We first check the easily proven formula
d(fg) = g df + f dg,
which follows from the Leibniz rule. Suppose that
ω =∑
i1<···<ik
fi1···ik dxi1 ∧ · · · ∧ dxik ,
φ =∑
i1<···<ik
gi1···ik dxi1 ∧ · · · ∧ dxik .
Then
ω ∧ φ =∑
i1<···<ik
∑j1<···<j`
fi1···ikgj1···j` dxi1 ∧ · · · ∧ dxik ∧ dxj1 ∧ · · · ∧ dxj` ,
and hence
d(ω ∧ φ) =∑∑
d(fi1···ikgj1···j`) dxi1 ∧ · · · ∧ dxik ∧ dxj1 ∧ · · · ∧ dxj`
=∑∑
(dfi1···ikgj1···j` + fi1···ikgj1···j`) dxi1 ∧ · · · ∧ dxik ∧ dxj1 ∧ · · · ∧ dxj`
=∑∑
dfi1···ik dxi1 ∧ · · · ∧ dxik ∧ gj1···j` dxj1 ∧ · · · ∧ dxj`
+∑∑
(−1)kfi1···ik dxi1 ∧ · · · ∧ dxik ∧ dgj1···j` ∧ dxj1 ∧ · · · ∧ dxj`
= dω ∧ φ+ (−1)k ω ∧ dφ.
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Axiom 2: We use equality of mixed partials. First we show that d(df) = 0 whenf ∈ A0(M):
d(df) = d
(n∑j=1
∂f
∂xjdxj
)
=n∑
i,j=1
∂2f
∂xj∂xidxj ∧ dxi
=∑j<i
(∂2f
∂xj∂xi− ∂2f
∂xi∂xj
)dxj ∧ dxi
= 0.
In general, if ω ∈ Ak(M), say ω =∑
i1<···<ik fi1···ik dxi1 ∧ · · · ∧ dxik ,
d dω =∑
i1<···<ik
d(dfi1···ik ∧ dxi1 ∧ · · · ∧ dxik)
=∑
i1<···<ik
d(dfi1···ik) ∧ dxi1 ∧ · · · ∧ dxik −∑
i1<···<ik
dfi1···ik d(dxi1 ∧ · · · ∧ dxik)
= 0.
This finishes local existence.
For global existence, cover M by coordinate systems (Uα, (x1α, . . . , x
nα)) : α ∈ A.
On each Uα we have locally defined maps
dα : Ak(Uα) → Ak+1(Uα).
If Uα ∩ Uβ 6= ∅, then
dα = dβ : Ak(Uα ∩ Uβ) → Ak+1(Uα ∩ Uβ).
Therefore the locally defined dα’s fit together to give global operators
d : Ak(M) → Ak+1(M).
This finishes the proof.
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15.1 Appendix: Universal Description of ΛkT ∗pM
We consider the multilinear map
π : T ∗pM × · · · × T ∗
pM︸ ︷︷ ︸k
→ ΛkT ∗pM
defined byπ(α1, . . . , αk) = α1 ∧ · · · ∧ αk.
Proposition. Given any alternating multilinear map T:T∗pM×· · ·×T ∗
pM → V , there
is a linear map T : ΛkT ∗pM → V which makes the following diagram commute:
T ∗pM × · · · × T ∗
pMT //
π ((QQQQQQQQQQQQV
ΛkT ∗pM
eT<<yyyyyyyyy
Proof. T must be given by the formula
T (dxi1 |p ∧ · · · ∧ dxik |p) = T (dxi1 |p, . . . , dxik |p).
This formula also proves existence.
79
Chapter 16
November 15, 2005
If F : M → N is a smooth map, we have a linear map F ast : Ak(N) → Ak(M) whichis defined by
F ∗(ω)(v1, . . . , vk) = ω(F∗(v1), . . . , F∗(vk))
for v1, . . . , vk ∈ TpM .
Proposition.
(a) F ∗(ω ∧ θ) = F ∗(ω) ∧ F ∗(θ).
(b) d(F ∗ω) = F ∗ dω.
Proof. (a) is an exercise. For (b), we first note that if f :∈ A0(M), we defineF ∗(f) = f F . Then
d(F ∗(f))(v) = df(F∗v) = F∗(v)(f)
d(f F )(v) = v(f F ) = F∗(v)(f),
so (b) holds when ω ∈ A0(M).
Next we check (b) for ω ∈ Ak(U), where U is the domain of local coordinates. If
ω =∑
i1<···<ik
fi1···ik dxi1 ∧ · · · ∧ dxik ,
dω =∑
i1<···<ik
dfi1···ik ∧ dxi1 ∧ · · · ∧ dxik ,
80
and hence
F ∗(dω) =∑
i1<···<ik
F ∗dfi1···ik ∧ F ∗dxi1 ∧ · · · ∧ F ∗dxik
=∑
i1<···<ik
d(fi1···ik F ) ∧ d(xi1 F ) ∧ · · · ∧ d(xik F ).
On the other hand,
F ∗(ω) =∑
i1<···<ik
(fi1···ik F ) d(xi1 F ) ∧ · · · ∧ d(xik F ),
and using the axioms for exterior derivative, we find that
d(f ∗(ω)) =∑
i1<···<ik
d(fi1···ik F ) ∧ d(xi1 F ) ∧ · · · ∧ d(xik F ).
Thus we see that F ∗(dω) = d(F ∗ω) on U . Since d and F ∗ are local operators,
F ∗ d = d F ∗ : Ak(N) → Ak+1(M).
Let Zk(M) = ω ∈ Ak(M) : dω = 0, the vector space of closed differential forms.Let Bk(M) = ω ∈ Ak(M) : ω = dθ for some θ ∈ Ak−1(M), the vector space ofexact differential forms. Since d d = 0, Bk(M) ⊆ Zk(M).
Definition. Hk(M) = Zk(M)/Bk(M) is the k-dimensional de Rham cohomologygroup of M .
(It is called a group even though it is more than that, a real vector space.)
The preceding proposition shows that whenever F : M → N is a smooth map, wehave a commutative ladder
· · · // Ak−1(N) //
F ∗
Ak(N) //
F ∗
Ak+1(N) //
F ∗
· · ·
· · · // Ak−1(M) // Ak(M) // Ak+1(M) // · · ·
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It follows from commutativity that
F ∗(Zk(N)) ⊆ Zk(M), F ∗(Bk(N)) ⊆ Bk(M),
and hence F ∗ induces a linear map
F ∗ : Hk(N) → Hk(M).
Let H∗(M) =∑n
k=0Hk(M).
If ω is a closed k-form representing [ω] ∈ Hk(M) and ϕ is a closed `-form representing[ϕ] ∈ H`(M), then ω ∧ ϕ is a closed (k + `)-form representing [ω ∧ ϕ] ∈ Hk+`(M).
Suppose ω and ω′ both represent the same element in Hk(M). Then ω − ω′ is exact,i.e. ω − ω′ = dθ for some θ ∈ Hk−1(M). Then d(θ ∧ ϕ) = dθ ∧ ϕ = ω ∧ ϕ − ω′ ∧ ϕ,so ω ∧ ϕ and ω′ ∧ ϕ represent the same element of Hk+`(M). Similarly, if ϕ and ϕ′
represent the same element of H`(M), ω∧ϕ and ω∧ϕ′ represent the same element ofHk+`(M). Thus if [ω] ∈ Hk(M) and [ϕ] ∈ H`(M), we can define the cup product
[ω] ^ [ϕ] = [ω ∧ ϕ] ∈ Hk+`(M).
The cup product maps H∗(M) into an algebra over R, called the de Rham coho-mology algebra.
The correspondences
M 7→ Hk(M)
(F : M → N) 7→ (F ∗ : A∗(N) → A∗(M)),
where A∗(M) =∑n
k=0Ak(M), is a “contravariant function” from the category ofsmooth manifolds and smooth maps to the category of algebras and algebra homo-morphisms. In other words,
1. id : M →M induces id : H∗(M) → H∗(M).
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2. a commutative diagram
ME //
F BBB
BBBB
B P
NG
>>~~~~~~~
induces a commutative diagram
H∗(M) H∗(P )E∗oo
G∗yyttttttttt
H∗(N)
F ∗
eeKKKKKKKKK
with arrows reversed.
Example. Suppose M = R2. A one-form ω = M dx+N dy is closed iff dω = 0. It isexact iff ω = df for some smooth function f . Thus the method of exact differentialsshows that H1(R2) ∼= 0.
On the other hand, if M = R2 − (0, 0), the one-form
ω =−y dx+ x dy
x2 + y2
is closed but not exact because its line integral around the unit circle is nonzero, asshown in calculus courses.
16.1 Orientation
Two charts (U, (x1, . . . , xn)) and (V, (y1, . . . , yn)) on a smooth manifold M are co-herently oriented if
det
(∂yi
∂xj
)> 0 on U ∩ V.
A smooth manifold M is orientable if it has an atlas of coherently oriented charts.Such an atlas defines an orientation for M . An oriented manifold is a manifoldwith a choice of orientation.
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A manifold is nonorientable if it does not possess an atlas of coherently orientedcharts.
Example. RP 2 is nonorientable.
SupposeM is a smooth manifold with orientation defined by the atlas (Uα, (x1α, . . . , x
nα)).
A new chart (V, (y1, . . . , yn)) is positively oriented if det(∂yi/∂xjα) > 0 for all αand negatively oriented if det(∂yi/∂xjα) < 0 for all α.
Exercise. A connected manifold M is orientable if and only if there is a nowherezero n-form Ω on M .
16.2 Local Integration of an n-Form
Suppose M is a smooth oriented n-dimensional smooth manifold, (U,ϕ) a positivelyoriented chart with ϕ = (x1, . . . , xn), and ω is a smooth n-form such that
supp(f) = closure of p ∈M : ω(p) 6= 0
compact and contained in U . Then
ω = f dx1 ∧ · · · ∧ dxn
for some function f , and we can define the integral∫M
ω =
∫U
ω =
∫ϕ(U)
(f ϕ−1) dx1 · · · dxn,
the last integral being the ordinary Riemann integral in Rn.
Is the integral well-defined?
84
Yes, by the change of variable formula from advanced calculus. (See, e.g. Rudin,Principles of Mathematical Analysis, 3rd edition, 1976, Theorem 10.9.)
Indeed, if (U, ψ) is a second smooth chart on the same open set U with ψ = (y1, . . . , yn),then
ω = f dx1 ∧ · · · ∧ dxn
= f
n∑i1=1
· · ·n∑
in=1
∂x1
∂yi1∂x2
∂yi2· · · ∂x
n
∂yindyi1 ∧ · · · ∧ dyin
= f∑σ∈Sn
sgnσ∂x1
∂yσ(1)· · · ∂xn
∂yσ(n)dy1 ∧ · · · ∧ dyn
= f det
(∂xi
∂yj
)dy1 ∧ · · · ∧ dyn.
If both (x1, . . . , xn) and (y1, . . . , yn) are positively oriented, det(∂xi/∂yj) > 0 andω = f | det(∂xi/∂yj)| dy1 ∧ · · · ∧ dyn. The change of variables formula simply says∫
ϕ(U)
(f ϕ−1) dx1 ∧ · · · ∧ dxn =
∫ψ(U)
(f ψ−1)
∣∣∣∣det
(∂xi
∂yj
)∣∣∣∣ ψ dy1 ∧ · · · ∧ dyn,
which implies that the integral is indeed well-defined.
85
Chapter 17
November 17, 2005
Suppose M is an oriented n-dimensional manifold and ω is a smooth n-form on Mwith compact support contained in U , where U is the domain of a positively orientedsmooth coordinate system ϕ = (x1, . . . , xn).
If ω = f dx1 ∧ · · · ∧ dxn, we define the integral of ω over U by∫U
ω =
∫ϕ(U)
(f ϕ−1)−1 dx1 · · · dxn,
the integral on the right being the Riemann or Lebesgue integral.
If ϕ = (y1, . . . , yn) is another positively oriented smooth coordinate system on U ,then
f dx1 ∧ · · · ∧ dxn = f det
(∂xi
∂yj
)dy1 ∧ · · · ∧ dyn
= f
∣∣∣∣det
(∂xi
∂yj
)∣∣∣∣ dy1 ∧ · · · ∧ dyn.
It follows from the change of variable formula for multiple integrals that∫ϕ(U)
f ϕ−1 dx1 · · · dxn =
∫ψ(U)
f ψ−1
∣∣∣∣det
(∂xi
∂yj
)∣∣∣∣ ψ−1 dy1 · · · dyn,
so∫Uω is independent of choice of positively oriented local coordinates.
86
Integration has the usual linearity property:∫U
(c1ω1 + c2ω2) = c1
∫U
ω1 + c2
∫U
ω2,
when c1 and c2 are constants.
17.1 Global Integration
If ω is an n-form with compact support on an oriented n-manifold Mn, we define∫Mω as follows:
First choose a finite collection (Uα, ϕα) : α ∈ A of positively oriented coordinatesystems such that suppω ⊆
⋃Uα : α ∈ A. Let ψα : α ∈ A be a partition of unity
subordinate to Uα : α ∈ A. Then define∫M
ω =∑α∈A
∫Uα
ψα · ω.
To see that the integral is well-defined, suppose that (Vβ, ξβ), β ∈ B is a secondfinite collection of positively oriented charts covering supp(ω). Let ηβ : β ∈ B be apartition of unity subordinate to Vβ : β ∈ B. Then∑
α∈A
∫Uα
ψαω =∑α∈A
∑β∈B
∫Uα∩Vβ
ψαηβω =∑β∈B
∫Vβ
ηβω.
Hence the integral is well-defined.
A “k-dimensional singular compact oriented parametrized manifold” in M (or morebriefly a “singular k-manifold”) is a pair (Nk, F ), where Nk is a smooth compactk-manifold and F : Nk →Mn is a smooth map.
87
An element ω ∈ Ak(M) is the integrand for the integral over a singular k-manifold,∫(Nk,F )
ω =
∫Nk
F ∗ω.
If F is the inclusion, we sometimes write∫Nk ω.
17.2 Manifolds with Boundary
Let Rn− = (x1, . . . , xn) ∈ Rn : x1 ≤ 0 with the subspace topology. ∂Rn
− =(x1, . . . , xn) ∈ Rn : x1 = 0. We will identify ∂Rn
0 with Rn−1.
Definition. An n-dimensional smooth manifold with boundary is a secondcountable Hausdorff space M together with a collection A = (Uα, ϕα) : α ∈ A suchthat
1. Each ϕα is a homeomorphism from the open set Uα ⊆ M to the open setϕ(Uα) ⊆ Rn
−.
2. ϕβ ϕ−1α is C∞ where defined.
3. Uα : α ∈ A cover M .
The boundary of M is
∂M = p ∈M : ϕα(p) ∈ ∂Rni for some α ∈ A.
Lemma. If ϕα(p) =∈ ∂Rn− for some α ∈ A, then ϕβ(p) ∈ ∂Rn
− for all β ∈ A forwhich ϕβ(p) is defined.
Proof. Suppose to the contrary that ϕβ(p) ∈ interior of Rn−. Note that ϕα ϕ−1
β
has a nonsingular derivative at ϕβ(p). Hence by the inverse function theory ϕα ϕ−1β
maps an open neighborhood Vα of ϕα(p) onto an open neighborhood Vβ of ϕβ(p).Vα is open in Rn and Vα ⊆ Rn
−; hence ϕα(p) is contained in the interior of Rn−, a
88
contradiction.
If α ∈ A, let Vα = Uα ∩ ∂M , ψα = ϕα |Vα . Then (Vα, ψα) : α ∈ A is a smooth atlason ∂M making ∂M into an (n− 1)-dimensional smooth manifold.
If Mn is a smooth n-manifold without boundary, we write ∂Mn = ∅. Note that∂(∂M) = ∅ whenever M is a smooth manifold with boundary.
Orientation. Suppose that Mn is an oriented manifold with boundary, i.e. the el-ements of A are coherently oriented. Thus if (U, (x1, . . . , xn)) and (V, (y1, . . . , yn))are two charts of A, det(∂yi/∂xj) > 0. Then U ∩ ∂M = p ∈ M : x1(p) = 0 andV ∩∂M = p ∈M : y1(p) = 0, and (U∩∂M, (x2, . . . , xn)) and (V ∩∂M, (y2, . . . , yn))are smooth charts for ∂M . We claim they are coherently oriented.
Indeed, if p ∈ ∂M ∩ (U ∩ V ), ∂y1
∂xi (p) = 0 for 2 ≤ i ≤ n since ∂∂xi
∣∣p
is tangent to ∂M
and y1 is constant along ∂M . On the other hand, ∂∂x1
∣∣p
points out of M , i.e. in the
direction of increasing y1, so ∂y1
∂x1 (p) > 0. Thus
det
∂y1
∂x1 (p) 0 · · · 0
0 ∂y2
∂x2 (p) · · · ∂y2
∂xn (p)...
......
0 ∂yn
∂x2 (p) · · · ∂yn
∂xn (p)
> 0 ⇒ det
∂y2
∂x2 (p) · · · ∂y2
∂xn (p)...
...∂yn
∂x2 (p) · · · ∂yn
∂xn (p)
> 0.
Thus (U∩∂M, (x2, . . . , xn)) and (V ∩∂M, (y2, . . . , yn)) are indeed coherently oriented.
Hence an orientation on a smooth manifold with boundary induces an orientation onits boundary.
Stokes’s Theorem. Let Mn be an oriented smooth n-manifold with boundary ∂Mn.Give ∂Mn the induced orientation, and let ι : ∂Mn →Mn denote the inclusion map.If θ is a smooth (n− 1)-form on M with compact support,∫
∂M
ι∗θ =
∫M
dθ.
89
Proof.
1. First cover M by positively oriented smooth coordinate systems (Uα, ϕα) : α ∈A such that if α ∈ A, either
(I) ϕα(Uα) = (a1α, b
1α)× · · · × (anα, b
nα) or
(II) ϕα(Uα) = (a1α, 0]× (a2
α, b2α)× · · · × (anα, b
nα).
Choose a partition of unity ψα : α ∈ A subordinate to Uα : α ∈ A.
2. It will suffice to prove Stokes’s Theorem for the special case that supp(θ) ⊆ Uαfor some α ∈ A. Indeed, assuming the special case, we find that if θ is anarbitrary (n− 1)-form with compact support, then∫
∂M
ι∗θ =∑α∈A
∫Uα∩∂M
ι∗(ψαθ)
=∑α∈A
∫Uα
α(ψαθ)
=
∫Uα
α
(∑α∈A
ψαθ
)= · · ·
=
∫M
dθ.
3. Thus it suffices to prove Stokes’s Theorem in the special case where (U,ϕ) is oftype (I) or (II). For simplicity, we consider type II only:
ϕ(U) = (a1, 0]× (a2, b2)× · · · × (an, bn).
The proof for type (I) is simpler. In fact we can assume ϕ is the identity. If θis a smooth (n− 1)-form with compact support contained in U , we can write
θ =n∑i=1
(−1)i−1fi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxn.
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Then if ι : Rn−1 → Rn− is the inclusion,
ι∗θ = f1 dx2 ∧ · · · ∧ dxn,
dθ =n∑i=1
∂fi∂xi
dx1 ∧ · · · ∧ dxn.
To verify Stokes’s Theorem, we need to calculate two integrals:∫U
dθ =
∫U
n∑i=1
∂f i
∂xidx1 · · · dxn
=n∑i=1
∫ bn
an
· · ·∫ b2
a2
∫ 0
a1
∂f i
∂xidx1 · · · dxn
=
∫ bn
an
· · ·∫ b2
a2
∫ a1
0
[f 1(0, x2, . . . , xn)− f 1(a1, x2, . . . , xn)] dx2 · · · dxn
+n∑i=2
∫ bn
an
· · ·∫ bi+1
ai+1
∫ bi−1
ai−1
· · ·∫ 0
a1
[f i(bi, x2, . . . , xn)− f i(ai, x2, . . . , xn)] dx1 · · · dxi−1dxi+1 · · · dxn
=
∫ bn
an
· · ·∫ b2
a2
f 1(0, x2, . . . , xn) dx1 · · · dxn.
On the other hand,∫∂U
ι∗θ =
∫∂U
(f 1 ι) dx2 ∧ · · · ∧ dxn
=
∫ bn
an
· · ·∫ b2
a2
(f 1 ι) dx2 · · · dxn
=
∫ bn
an
· · ·∫ b2
a2
f 1(0, x2, . . . , xn) dx2 · · · dxn.
The two integrals are equal, so Stokes’s Theorem is verified.
91
Chapter 18
November 22 2005
Recall that de Rham cohomology is defined as follows: We start with the cochaincomplex
· · · −−−→ Ak−1 d−−−→ Ak(M)d−−−→ Ak+1(M) −−−→ · · · ,
and let
Zk(M) = ker(d : Ak(M) → Ak+1(M)),
Bk(M) = im(d : Ak−1(M) → Ak(M)).
Then d d = 0 implies that Bk(M) ⊆ Zk(M), and we set
Hk(M) =Zk(M)
Bk(M).
If ω is a k-form such that dω = 0, we let [ω] denote its corresponding equivalenceclass in Hk(M).
We let H∗(M) =∑n
k=0Hk(M) and define the cup product on Hk(M) by [ω] ^
[ϕ] = [ω ∧ ϕ]. H∗(M) is an algebra over R.
A smooth map F : M → N induces an algebra homomorphism F ∗ : H(N) → H∗(M).We call H∗(M) the de Rham cohomology algebra of M . It is a basic topologicalinvariant of M .
To calculate H∗(M), we need
92
(a) the Homotopy Theorem.
(b) the Mayer-Vietoris sequence.
Suppose F,G : M → N are smooth maps. A smooth homotopy from F to G is asmooth map
H : M × [0, 1] → N
such that H(p, 0) = F (p) and H(p, 1) = G(p). If such a homotopy exists, we say thatF and G are smoothly homotopic, and we write F ' G.
Homotopy Theorem. If F,G : M → N are smoothly homotopic, then F ∗ = G∗ :H∗(N) → H∗(M).
Corollary. (Poincare Lemma) If U is a convex open subset of Rn, then
Hk(U) ∼=
R if k = 0,
0 otherwise.
Proof. We first note that a point p0 can be regarded as a 0-dimensional manifold,and
Hk(p0) ∼=
R if k = 0,
0 otherwise.
Let p0 ∈ U , and let F : p0 →M be the inclusion, G : M → p0 the constant map.Then G F = id. We claim that F G ' id. To show this we define
H : U × [0, 1] → U by H(p, t) = tp+ (1− t)p.
Then H(p, 0) = F G(p), while H(p, 1) = id. Thus G F = id, so F ∗ G∗ = idand F G ' id, so G∗ F ∗ = id, from which the Poincare Lemma is an immediateconsequence.
To prove the Homotopy Theorem, we reduce it to simpler statements easier to prove:First we reduce the Homotopy Theorem to the
93
Cylinder Lemma. Suppose that J0, J1 : M →M × [0, 1] are defined by
J0(p) = (p, 0), J1(p) = (p, 1).
Let J∗0 , J∗1 : Hk(M × [0, 1]) → Hk(M) be the induced maps on cohomology. Then
J∗0 = J∗1 .
Proof that Cylinder Lemma ⇒ Homotopy Theorem. If F,G : M → N aresmoothly homotopic, we have a commutative diagram
MF
''J0 %%JJJJJJJJJJ
M × [0, 1]H // N
M
J1
99tttttttttt G
77
Hence F ∗ = J∗0 H∗ = J∗1 H∗ = G∗.
The Cylinder Lemma in turn is based upon
Basic Lemma. Suppose that J∗0 , J∗1 : Ak(M × [0, 1]) → Ak(M) are the induced
maps on k-forms. Then there exist linear maps
∆ : Ak(M × [0, 1]) → Ak−1(M)
such thatJ∗1 − J∗0 = ∆ d+ d ∆.
Proof that Basic Lemma ⇒ Cylinder Lemma. Suppose ω ∈ Zk(M × [0, 1])with cohomology class [ω] ∈ Hk(M × [0, 1]). Then
J∗1ω − J∗0ω = d∆ω,
94
and hence [J∗0ω] = [J∗1ω].
Proof of Basic Lemma. Let (Uα, (x1α, . . . , x
nα)) : α ∈ A be a smooth atlas on M .
It suffices to show that there exist maps
∆α : Ak(Uα × [0, 1]) → Ak−1(Uα)
such that
(a) J∗1 − J∗0 = ∆α d+ d ∆α on Ak(Uα × [0, 1]),
1. ∆α = ∆β on Ak((Uα ∩ Uβ)× [0, 1]).
Let t : [0, 1] → R be the coordinate defined by t(s) = s. Then (t, x1α, . . . , x
nα) is a
coordinate system on Uα × [0, 1]. Any k-form on Uα × [0, 1] is a linear combinationof monomials of the form
I. f(t, x1α, . . . , x
nα) dx
i1α ∧ · · · ∧ dxikα ,
II. f(t, x1α, . . . , x
nα) dt ∧ dxi1α ∧ · · · ∧ dx
ik−1α .
We define ∆α by∆α(f(t, x1, . . . , xn) dxi1 ∧ · · · ∧ dxik) = 0,
∆α(f(t, x1, . . . , xn) dt ∧ dxi1 ∧ · · · ∧ dxik−1)
=
[∫ 1
0
f(u, x1, . . . , xn) du
]dxi1 ∧ · · · ∧ dxik−1 .
Case I. We need to show that
J∗1 (f dxi1 ∧ · · · ∧ dxik)− J∗0 (f dxi1 ∧ · · · ∧ dxik)= ∆d(f dxi1 ∧ · · · ∧ dxik) + d∆(f dxi1 ∧ · · · ∧ dxik).
95
But d∆(f dxi1 ∧ · · · ∧ dxik) = 0 and
∆d(f dxi1 ∧ · · · ∧ dxik) = ∆
(∂f
∂tdxi1 ∧ · · · ∧ dxik
)=
[∫ 1
0
∂f
∂t(u, x1, . . . , xn)
]dxi1 ∧ · · · ∧ dxik
= f(1, x1, . . . , xn) dxi1 ∧ · · · ∧ dxik − f(0, x1, . . . , xn) dxi1 ∧ · · · ∧ dxik
= J∗1 (f dxi1 ∧ · · · ∧ dxik)− J∗0 (f dxi1 ∧ · · · ∧ dxik).
Case II. Since J∗1 dt = 0 = J∗0 dt,
J∗1 (f dt ∧ dxi1 ∧ · · · ∧ dxik−1) = 0 = J∗0 (f dt ∧ dxi1 ∧ · · · ∧ dxik−1).
Moreover,
d(f dt ∧ dxi1 ∧ · · · ∧ dxik−1) =n∑j=1
∂f
∂xjdxj ∧ dt ∧ dxi1 ∧ · · · ∧ dxik−1 ,
∆d(f dt ∧ dxi1 ∧ · · · ∧ dxik)
= −n∑j=1
[∫ 1
0
∂f
∂xj(u, x1, . . . , xn) du
]dxj ∧ dxi1 ∧ · · · ∧ dxik−1 .
The minus sign comes from interchanging dt and dxj.
On the other hand,
∆(f dt ∧ dxi1 ∧ · · · ∧ dxik−1) =
[∫ 1
0
f(u, x1, . . . , xn) du
]dxi1 ∧ · · · ∧ dxik−1 ,
d∆(f dt ∧ dxi1 ∧ · · · ∧ dxik−1)
=n∑j=1
[∫ 1
0
∂f
∂xj(u, x1, . . . , xn) du
]dxj ∧ dxi1 ∧ · · · ∧ dxik−1 .
96
Thus(∆d+ d∆)(f dt ∧ dxi1 ∧ · · · ∧ dxik−1) = 0.
Manifolds M and N are smoothly homotopy equivalent if there are smooth maps
F : M → N, G : N →M
such that G F ' id and F G ' id. It then follows from the Homotopy Theoremthat
G∗ : Hk(M) → Hk(N)
is an isomorphism.
For example, the circle is smoothly homotopy equivalent to the cylinder.
The second important computational tool is the Mayer-Vietoris sequence. Supposethat M = U ∪ V , where U and V are open subsets. We then have a commutativediagram
UiU
AAA
AAAA
A
U ∩ V
jU
;;wwwwwwwww
jV ##GGG
GGGG
GGM
ViV
>>
We construct a sequence of linear maps
0 −−−→ Ak(M)i∗−−−→ Ak(U)⊕Ak(V )
j∗−−−→ Ak(U ∩ V ) −−−→ 0 (∗)
by settingi∗ω = (i∗Uω, i
∗V ω), j∗(φU , φV ) = j∗U(φU)− j∗V (φV ).
Lemma. The sequence (∗) is exact.
97
To prove this we need to show first that i∗ is injective. But i∗(ω) = (i∗Uω, i∗V ω) = (0, 0),
so ω = 0.
We need to verify that j∗ i∗ = 0. But
j∗ i∗(ω) = j∗U i∗U − j∗V i
∗V ω = ω |U∩V −ω |U∩V = 0.
Next, we need to verify that if j∗(φU , φV ) = 0, then (φU , φV ) ∈ im i∗. But j∗(φU , φV ) =0 implies that φU |U∩V = φV |U∩V implies that there exists a φ ∈ Ak(M) such thati∗(φ) = (φU , φV ).
Finally, we need to verify that j∗ is surjective. For this, we choose a partition of unityψU , ψV subordinate to U, V . If θ ∈ Ak(U ∩ V ), define φU ∈ Ak(U) by
φU(p) =
ψV (p)θ(p) for p ∈ U ∩ V ,
0 for p ∈ U − (U ∩ V ).
φV (p) =
−ψU(p)θ(p) for p ∈ U ∩ V ,
0 for p ∈ V − (U ∩ V ).
Then j∗(φU , φV ) = θ.
98
Chapter 19
November 29, 2005
Recall that last time we considered a decomposition of a manifold M into two opensets, M = U ∪ V , which yields a commutative diagram of inclusions
UiU
AAA
AAAA
A
U ∩ V
jU
;;wwwwwwwww
jV ##GGG
GGGG
GGM
ViV
>>
This yields a short exact sequence
0 −−−→ Ak(M)i∗−−−→ Ak(U)⊕Ak(V )
j∗−−−→ Ak(U ∩ V ) −−−→ 0,
where i∗(ω) = (i∗Uω, i∗V ω) and j∗(φU , φV ) = j∗U(φU) − j∗V (φV ). This yields a commu-
tative latter in which the rows are exact:
0 // Ak(M)
i∗ //
d
Ak(U)⊕Ak(V )j∗ //
d
Ak(U ∩ V ) //
d
0
0 // Ak+1(M)i∗//
Ak+1(U)⊕Ak+1(V )j∗ //
Ak+1(U ∩ V ) //
0
in which the rows are exact. The commuting diagram gives maps
i∗ : Hk(M) → Hk(U)⊕Hk(V ),
jast : Hk(U)⊕Hk(V ) → Hk(U ∩ V ).
99
It also enables us to define a connecting homomorphism
∆ : Hk(U ∩ V ) → Hk+1(M).
The definition goes like this: If [θ] ∈ Hk(U ∩ V ), choose a representative θ ∈ Zk(U ∩V ) ⊆ Ak(U ∩ V ). Since j∗ is surjective, we can choose φ ∈ Ak(U)⊕Ak(V ) such thatj∗(φ) = θ. Then j∗(dφ) = dj∗φ = dθ = 0, so there is a unique ω ∈ Ak+1(M) suchthat i∗(ω) = dφ. (ω is unique because i∗ is injective.) Finally, i∗(dω) = di∗(ω) =d d(φ) = 0, and since i∗ is injective we conclude that dω = 0. We set
∆([θ]) = [ω].
Roughly speaking,∆ = (i∗)
−1 d (j∗)−1.
We need to check that ∆ is well-defined, i.e. that it is independent of the variouschoices made. We check, for example, that ∆([θ]) does not depend on the choice of
φ ∈ Ak(U)⊕Ak(V ). If φ and φ satisfy j∗(φ) = j∗(φ) = θ, then there is an ω ∈ Ak(M)
such that i∗ω = φ− φ. Then i∗(dω) = di∗ω = d(φ− φ) and dω = i−1∗ (dφ)− i−1
∗ (dφ), so
[i−1∗ (dφ)] = [i−1
∗ (dφ)], so indeed ∆([θ]) does not depend on choice of φ. Similarly oneshows that ∆([θ]) does not depend on the choice of representative θ ∈ [θ]. Clearly, ∆is a linear map.
Theorem. (Mayer-Vietoris sequence) If M = U ∩V , the following sequence is exact:
· · ·
rreeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Hk(M)i∗ // Hk(U)⊕Hk(V ) // Hk(U ∩ V )
rreeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Hk+1(M)i∗// Hk+1(U)⊕Hk+1(V ) // Hk+1(U ∩ V )
rreeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
· · ·
In other words, the image of each homomorphism is the kernel of the next.
For example, we check that im j∗ = ker ∆.
100
im(j∗) ⊆ ker ∆. Suppose [θ] ∈ im j∗, so [θ] = j∗([φ]) for some [φ] ∈ Hk(U)⊕Hk(V ).Then there are representatives θ for [θ] and φ for [φ] such that j∗φ = θ. Note thatdφ = 0. Hence ∆([θ]) = [i−1
∗ d j−1∗ (θ)] = [i−1
∗ (dφ)] = 0, so [θ] ∈ ker ∆.
ker(∆) ⊆ im(j∗). Suppose [θ] ∈ ker ∆. Then i−1∗ dj−1
∗ (θ) is exact, i.e. if φ ∈ j−1∗ (θ),
i−1∗ (dφ) = dω for some ω ∈ Ak(M). But then d(φ − i∗ω) = 0 and j∗(φ − i∗ω) = θ.
Thus [θ] = j∗([φ− i∗ω]), and [θ] ∈ im j∗.
The other implications are quite similar.
We now compute Hk(S1). Let U and V be arcs of the circle covering the entire circle.Then
Hk(U) ∼= Hk(V ) '
R if k = 0,
0 if k 6= 0,Hk(U ∩ V ) ∼=
R⊕ R if k = 0,
0 if k 6= 0.
H0(S1) = f : S1 → R | df = 0, the set of constant functions on S1. Hence theMayer-Vietoris sequence starts out as
0
ttjjjjjjjjjjjjjjjjjjjjjj
R // R⊕ R // R
uujjjjjjjjjjjjjjjjjjj
H1(S1) // 0 // 0
uukkkkkkkkkkkkkkkkkkk
H2(S1) // 0 // 0
uujjjjjjjjjjjjjjjjjjjjjj
0
Thus Hk(S1) ∼=
R if k = 0 or 1,
0 if k 6= 0 or 1.
101
At the next stage, we see that S2 = U ∪ V , where
Hk(U) =∼= Hk(V ) '
R if k = 0,
0 if k 6= 0,
Hk(U ∩ V ) ∼= Hk(S1) ∼=
R if k = 0 or 1,
0 if k 6= 0 or 1.
The Mayer-Vietoris sequence yields
0
ttjjjjjjjjjjjjjjjjjjjjjj
R // R⊕ R // R
uujjjjjjjjjjjjjjjjjjj
H1(S2) // 0 // R
uukkkkkkkkkkkkkkkkkkk
H2(S2) // 0 // 0
ttjjjjjjjjjjjjjjjjjjjjj
· · ·
This time we conclude that
Hk(S2) ∼=
R if k = 0 or 2,
0 otherwise.
By induction, we establish that
Hk(Sn) ∼=
R if k = 0 or n,
0 otherwise.
Proposition. If M is a compact oriented manifold of dimension n, Hn(M) 6= 0.
Proof. Cover M by positively oriented charts (Uα, (x1α, . . . , x
nα)) : α ∈ A and
let ψα : α ∈ A be a partition of unity subordinate to Uα : α ∈ A. Let
102
ωα = ψα dx1α ∧ · · · ∧ dxnα, ω =
∑ψαωα. Then
∫Mω 6= 0, so by Stokes’s Theo-
rem, ω 6= dθ for any (n− 1)-form θ. ω is closed but not exact, so Hn(Mn) 6= 0.
Proposition. H0(M) = R⊕ · · · ⊕ R︸ ︷︷ ︸k
, where k is the number of components of M .
Proof. Z0(M) = f : M → R | df = 0, the set of locally constant functions,while B0(M) = 0. A locally constant function can take on any real value on eachcomponent. Hence H0(M) ' R⊕ · · ·R︸ ︷︷ ︸
k
, where k is the number of components of M .
We now let M be the union of two cylinders along a square region U ∩ V . In thiscase, the Mayer-Vietoris sequence gives
0
ttiiiiiiiiiiiiiiiiiiiiii
R // R⊕ R // R
uujjjjjjjjjjjjjjjjjjj
H1(M) // R⊕ R // 0
uukkkkkkkkkkkkkkkkkkkk
H2(M) // 0 // · · ·
We conclude that
Hk(M) ∼=
R if k = 0,
R⊕ R if k = 1,
0 otherwise.
We next consider the torus T 2 which contains a disc D2 containing a smaller disc D1.Let U = T 2−closure of D1 and V =interior of D2, so U ∩ V is homeomorphic to M .
103
The Mayer-Vietoris sequence in this case yields
0
ttjjjjjjjjjjjjjjjjjjjjjj
Ri0 // R⊕ R i1 // R
i2
uujjjjjjjjjjjjjjjjjjj
H1(T 2)i3 // R⊕ R i4 // R
i5
uukkkkkkkkkkkkkkkkkkk
H2(T 2) // 0
Since H2(T 2) 6= 0, i5 must be an isomorphism. Hence
Hk(T 2) ∼=
R if k = 0 or 2,
R⊕ R if k = 1,
0 otherwise.
Let Σ2 be the sphere with two handles. We can decompose in the natural way anduse the Mayer-Vietoris sequence to show that
Hk(Σ2) ∼=
R if k = 0 or 2,
R⊕ R⊕ R⊕ R if k = 1,
0 otherwise.
Exercise. Carry this argument out.
By induction, one shows that if Σg is the sphere with g handles, then
Hk(Σg) ∼=
R if k = 0 or 2,
R2g if k = 1,
0 otherwise.
104
Application. Let Dn = (x1, . . . , xn) ∈ Rn | (x1)2+ · · ·+(xn)2 ≤ 1. Dn is a smoothmanifold with boundary, ∂Dn = Sn−1.
No Retraction Theorem. If n ≥ 2 there does not exist a smooth map F : Dn → Dn
which leaves every point of Sn−1 fixed.
Proof. Suppose F were such a map. Let x = F (0). Define H : Sn−1 × [0, 1] → Sn−1
byH((x1, . . . , xn), t) = F (tx1, . . . , txn).
H is a homotopy from the constant map F : Sn−1 → x to the identity map G.But then F ∗ : Hn−1(Sn−1) → Hn−1(Sn−1) is the zero map, while G∗ : Hn−1(Sn−1) →Hn−1(Sn−1) is the identity. Since Hn−1(Sn−1) ∼= R, this gives a contradiction.
Lemma. A smooth map G : Dn → Dn must have a fixed point.
Proof. If G has no fixed point, define F : Dn → Sn−1 as follows: Let L(x) bethe line through x and G(x), and let F (x) be the point on Sn−1 ∩ L(x) closest tox. F : Dn → Sn−1 is smooth and leaves Sn−1 pointwise fixed, contradicting the NoRetraction Theorem.
Brouwer Fixed Point Theorem. A continuous map F : Dn → Dn must have atleast one fixed point.
Proof. Suppose F : Dn → Dn has no fixed point. Let µ = inf|f(x)− x| : x ∈ Dn,µ > 0. Choose a smooth map G : Dn → Dn such that |F (x)−G(x)| < 1
2µ for x ∈ Dn.
Then G contradicts the previous lemma.
To get the smooth approximation, we can use the Weierstraß approximation theoremfrom real analysis. This theorem implies that there exists a polynomial P : Dn → Rn
which approximates F as closely as desired, |P (x) − F (x)| < ε, where ε > 0 isarbitrarily small. Let G = 1
1+εP . Then G : Dn → Dn is a smooth map with
|G(x)− F (x)| < ε+ ε1+ε
< 2ε.
105
Chapter 20
December 1, 2005
A pseudo-Riemannian metric 〈 , 〉 on a smooth manifold M determines isomorphisms
[ : TpM → T ∗pM, ] : T ap stM → TpM
by[(v)(w) = 〈v, w〉, ] = ([)−1.
Nondegeneracy of 〈 , 〉 implies that [ and ] are isomorphisms.
In local coordinates, 〈 , 〉 =∑n
i,j=1 gij dxi ⊗ dxj and
[
(∂
∂xi
∣∣∣∣p
)=
n∑j=1
gij(p) dxj |p,
](dxi |p) =n∑j=1
gij(p)∂
∂xj
∣∣∣∣p
,
where (gij(p)) is the matrix inverse to (gij(p)).
A pseudo-Riemannian metric 〈 , 〉 on M determines a nondegenerate symmetricbilinear form 〈 , 〉 on T ∗
pM by
〈α, β〉 = 〈]α, ]β〉, for α, β ∈ T ∗pM.
106
Define a multilinear map
µ : T ∗pM × · · · × T ∗
pM︸ ︷︷ ︸ k × T ∗pM × · · · × T ∗
pM︸ ︷︷ ︸k
→ R
by
µ((α1, . . . , αk), (β1, . . . , βk)) = det
〈α1, β1〉 · · · 〈α1, βk〉...
...〈αk, β1〉 · · · 〈αk, βk〉
.
This multilinear map is alternating in each set of variables, hence (see Appendix toLecture 15) induces a bilinear map
〈 , 〉 : ΛkT ∗pM × ΛkT ∗
pM → R
such that
〈α1 ∧ · · · ∧ αk, β1 ∧ · · · ∧ βk〉 = det
〈α1, β1〉 · · · 〈α1, βk〉...
...〈αk, β1〉 · · · 〈αk, βk〉
.
Let (θ1, . . . , θn) be an orthonormal basis for T ∗pM . By this we mean that
〈θi, θj〉 =
±1 if i = j,
0 if i 6= j.
Then
〈θi1 ∧ · · · ∧ θik , θj1 ∧ · · · ∧ θjk〉 =
±1 if (i1, . . . , ik) is a permutation of (j1, . . . , jk),
0 otherwise.
Henceθi1 ∧ · · · ∧ θik : 1 ≤ i1 < i2 < · · · < ik ≤ n
is an orthonormal basis for ΛkT ∗pM .
107
Special case: ΛnT ∗pM is one-dimensional and has an orthonormal basis consisting of
one elementθ1 ∧ · · · ∧ θn.
Suppose that M is oriented. We say that an orthonormal basis (θ1, . . . , θn) is posi-tively oriented if θ1∧ · · · ∧ θn is a positive multiple of dx1 |p ∧ · · · ∧ dxn |p wheneverthe coordinates (x1, . . . , xn) are positively oriented. Let
Θ = θ1 ∧ · · · ∧ θn,
whenever (θ1, . . . , θn) is positively oriented.
We define a linear isomorphism, the so-called Hodge star,
∗ : ΛkT ∗pM → Λn−kT ∗
pM
byα ∧ ∗β = 〈α, β〉Θ.
If (θ1, . . . , θn) is a positively oriented orthonormal basis for T ∗pM ,
∗(θi1 ∧ · · · ∧ θik) = 〈θi1 ∧ · · · ∧ θik , θi1 ∧ · · · ∧ θik〉θj1 ∧ · · · ∧ θjn−k ,
when (i1, . . . , ik, j1, . . . , jn−k) is a positive permutation of (1, . . . , n).
Note that if α ∈ ΛkT ∗pM ,
∗ ∗ α = (−1)k(n−k)α.
The Hodge star is also defined on differential forms
∗ : Ak(M) → An−k(M),
by (∗ω)(p) = ∗(ω(p)). Note that ∗1 = θ1 ∧ · · · ∧ θn = Θ. We call Θ the volumeform. If Mn is a compact Riemannian manifold,
volume of Mn =
∫Mn
Θ.
Examples.
108
1. R2 with coordinates (x, y) and metric
〈 , 〉 = dx⊗ dx+ dy ⊗ dy.
∗dx = dy, ∗dy = −dx, ∗(M dx + N dy) = −N dy +M dy. ∗ can be regardedas counterclockwise rotation through 90. Note that
(d ∗ d)(f) = d ∗(∂f
∂xdx+
∂f
∂ydy
)= d
(−∂f∂y
dx+∂f
∂xdy
)= −∂
2f
∂y2dy ∧ dx+
∂2f
∂x2dx ∧ dy
=
(∂2f
∂x2+∂2f
∂y2
)dx ∧ dy.
2. R3 with coordinates (x, y, z) and metric
〈 , 〉 = dx⊗ dx+ dy ⊗ dy + dz ⊗ dz.
In this case,
∗dx = dy ∧ dz,∗dy = dz ∧ dx,∗dz = dx ∧ dy,
∗(dy ∧ dz) = dx,
∗(dz ∧ dx) = dy,
∗(dx ∧ dy) = dz,
∗(P dx+Q dy +R dz) = P dy ∧ dz +Q dz ∧ dx+R dx ∧ dy,
d ∗ d(P dx+Q dy +R dz) =
(∂2P
∂x2+∂2Q
∂y2+∂2R
∂z2
)dx ∧ dy ∧ dz.
If ω = Ax dx+ Ay dy + AZ dz, φ = Bx dx+By dy +Bz dz,
ω ∧ φ =
∣∣∣∣Ay AzBy Bz
∣∣∣∣ dy ∧ dz +
∣∣∣∣Az AxBz Bx
∣∣∣∣ dz ∧ dx+
∣∣∣∣Ax AyBx By
∣∣∣∣ dx ∧ dy,∗(ω ∧ ϕ) =
∣∣∣∣Ay AzBy Bz
∣∣∣∣ dx+
∣∣∣∣Az AxBz Bx
∣∣∣∣ dy +
∣∣∣∣Ax AyBx By
∣∣∣∣ dz,the “cross product of ω and φ.”
109
3. R4 with coordinates (t, x, y, z) and
〈 , 〉 = −dt⊗ dt+ dx⊗ dx+ dy ⊗ dy + dz ⊗ dz.
Then
∗(dt ∧ dx) = −dy ∧ dz,∗(dt ∧ dy) = −dz ∧ dx,∗(dt ∧ dz) = −dx ∧ dy,∗(dy ∧ dz) = dt ∧ dx,∗(dz ∧ dx) = dt ∧ dy,∗(dx ∧ dy) = dt ∧ dz.
Electricity and MagnetismElectric field E = (Ex, Ey, Ez), magnetic field B = (Bx, By, Bz). Maxwell’s equations(with speed of light c = 1:)
∇ ·B = 0
∇× E + ∂B∂t
= 0
∇ · E = 0
∇×B− ∂E∂t
= 0
no charge density, no current density.
In the space-time formulation, we replace E and B by the Faraday tensor
F = −Ex dt∧ dx−Ey dt∧ dy−Ez dt∧ dz+Bx dy ∧ dz+By dz ∧ dx+Bz ∧ dx∧ dy.
Then
∗F = Bx dt ∧ dx+By dt ∧ dy +Bz dt ∧ dz + Ex dy ∧ dz + Ey dz ∧ dx+ Ez dx ∧ dy.
110
Then
dF = −∂Ex∂y
dy ∧ dt ∧ dx− ∂Ex∂z
dz ∧ dt ∧ dx
− ∂Ey∂x
dx ∧ dt ∧ dy − ∂Ey∂z
dz ∧ dt ∧ dy
− ∂Ez∂x
dx ∧ dt ∧ dz − ∂Ez∂y
dy ∧ dt ∧ dz
+∂Bx
∂tdt ∧ dy ∧ dz +
∂Bx
∂xdx ∧ dy ∧ dz
+∂By
∂tdt ∧ dz ∧ dz +
∂By
∂ydy ∧ dz ∧ dx
+∂Bz
∂tdt ∧ dx ∧ dy +
∂Bz
∂zdy ∧ dz ∧ dx
=
(∂Bx
∂x+∂By
∂y+∂Bz
∂z
)dx ∧ dy ∧ dz
+
(∂Bx
∂t+∂Ez∂y
− ∂Ey∂z
)dt ∧ dy ∧ dz
+
(∂By
∂t+∂Ex∂z
− ∂Ez∂x
)dt ∧ dz ∧ dx
+
(∂Bz
∂t+∂Ey∂x
− ∂Ex∂y
)dt ∧ dx ∧ dy,
so
dF = 0 ⇐⇒
∇ ·B = 0∂B∂t
+∇× E = 0.
Similarly,
d ∗ F = 0 ⇐⇒
∇ · E = 0
∇×B− ∂E∂t
= 0.
Maxwell’s equations become dF = 0, d ∗ F = 0.
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Chapter 21
Problems
1. Let Gk(Rn) denote the set of k-dimensional linear subspaces of Rn. Show thatGk(Rn) is a smooth manifold. You can use the following steps if you want:
(a) Let V ∈ Gk(Rn) and let pV : Rn → V denote the orthogonal projection
from Rn to V . Let UV = V ∈ Gk(Rn) : pV maps V isomorphically into V .Let e1, . . . , en be an orthogonal basis for Rn such that e1, . . . , ek is an or-thonormal basis for V . If V ∈ UV show there is a unique basis v1, . . . , vkfor V such that
v1 = e1 + x1,k+1ek+1 + · · ·+ x1nen,
...
vk = ek + xk,k+1ek+1 + · · ·+ xknen.
(b) Let ϕV : UV → Rk(n−k) by ϕV (V ) =
x1,k+1 · · · x1n...
. . ....
xk,k+1 · · · xkn
. If V1, V2 ∈
Gk(Rn) show that ϕV1 ϕ−1V2
is C∞, where defined.
(c) Show that there is a unique topology on Gk(Rn) which makes each suchϕ a homeomorphism. Show that the topology is second countable andHausdorff.
2. Define F : P n(R) →M(n+ 1,R) = (n+ 1)× (n+ 1) real matrices by
F ([x1, . . . , xn+1]) =1
(x1)2 + · · ·+ (xn+1)2
(x1)2 · · · x1xn+1
.... . .
...xn+1x1 · · · (xn+1)2
.
112
In vector notation, F ([~x]) = ~xt~x|~x|2 . Show that
(a) F is an immersion.
(b) F is one-to-one.
(c) P n(R) is compact.
(d) F is an imbedding.
3. Let M = R2 with the usual coordinates (x, y) and let X = x ∂∂x
+ (3x − y) ∂∂y
.Show that X is complete and find the corresponding one-parameter group ofdiffeomorphisms corresponding to X.
4. Let M be a smooth manifold and X a smooth vector field on M such that
support of X = p ∈M : X(p) 6= 0 is compact.
Show that X is complete.
5. Let M = R3 with the usual coordinates (x, y, z) and let
X = −z ∂∂y
+ y∂
∂zY = −x ∂
∂z+ z
∂
∂xZ = −y ∂
∂x+ x
∂
∂y.
(a) Find the one-parameter groups ϕt : t ∈ R corresponding to X, Y , andZ.
(b) Calculate the Lie brackets [X,Y ], [Y, Z], [Z,X]. Recall that
[X,Y ](f) = X(Y (f))− Y (X(f)).
6. If X and Y are smooth vector fields on M and ω is a smooth one-form, showthat
dω(X, Y ) = X(ω(Y ))− Y (ω(X))− ω([X, Y ]).
7. Let
J =
−1 0 0 00 1 0 00 0 1 00 0 0 1
.
Let O(1, 3) = A ∈ GL(4,R) : ATJA = J. Let Sym(4) = 4 × 4 symmetricmatrices, and define F : GL(4,R) → Sym(4) by F (A) = ATJA, so O(1, 3) =F−1(J).
113
(a) Show that J is a regular value of F .
(b) Show that O(1, 3) is a submanifold of GL(4,R).
8. On R3 − 0, consider the differential form
ω =x dy ∧ dz + y dz ∧ dx+ z dx ∧ dy
(x2 + y2 + z2)3/2.
(a) Show that ω is closed, i.e. that dω = 0.
(b) Calculate∫S2 ω, where S2 = x2 + y2 + z2 = 1. Use the result to show
that ω is not exact.
9. Let M =torus−disk. Suppose we know that Hk(M) ∼=
R if k = 0
R⊕ R if k = 1
0 otherwise.
Use the Mayer-Vietoris sequence to determine Hk(Σ2), where Σ2 is the spherewith two handles.
114
