Calculus on ManifoldsSolution of Exercise Problems

Yan Zeng

Version 1.0, last revised on 2000-01-10.

AbstractThis is a solution manual of selected exercise problems from Calculus on manifolds: A modern

approach to classical theorems of advanced calculus, by Michael Spivak. If you would like to correct anytypos/errors, please send email to zypublic@hotmail.com.

Contents1 Functions on Euclidean Space 2

1.1 Norm and Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Subsets of Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Functions and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Differentiation 32.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Basic Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.4 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.5 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.6 Implicit Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Integration 63.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2 Measure Zero and Content Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.3 Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.4 Fubini’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.5 Partitions of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.6 Change of Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Integration on Chains 114.1 Algebraic Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.2 Fields and Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.3 Geometric Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.4 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Integration on Manifolds 175.1 Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.2 Fields and Forms on Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.3 Stokes’ Theorem on Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.4 The Volume Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.5 The Classical Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1

1 Functions on Euclidean Space1.1 Norm and Inner ProductI 1-2.

Proof. Re-examine the proof of Theorem 1-1(2), we see equality holds if and only if∑i xiyi = |

∑i xiyi| =

|x||y|. The second equality requires x and y are linearly dependent. The first equality requires ⟨x, y⟩ ≥ 0.Combined, we conclude x and y are linearly dependent and point to the same direction.

I 1-7.

Proof. (a) Theorem 1-2 (4) and (5) establish a one-to-one correspondence between norm and inner product.(b) If Tx = 0, |x| = |Tx| = 0, which implies x = 0. So T is injective. This further implies T is surjective

(Lax [1] page 15). So T is an isomorphism and T−1 is well-defined. It’s clear that T−1 is also norm preservingand hence must enjoy the same properties.

I 1-8.

Proof. Refer to [1]

I 1-9.

Proof. Use matrix to prove T is norm preserving. This will imply T is inner product preserving and henceangle preserving. Also use matrix to check ⟨x, Tx⟩ = cos θ|x||Tx|.

I 1-12.

Proof. Lax [1] page 66, Corollary 4′.

1.2 Subsets of Euclidean SpaceI 1-17.

Proof. Step1: We divide the square [0, 1]× [0, 1] into four equal squares by connecting ( 12 , 0) and (0, 12 ), (0,12 )

and (1, 12 ). We place on point in each of the squares and make sure no two points are on the same horizontalor vertical line.

· · ·Step n: We divide each of the squares obtained in Step (n-1) into four equal squares. We place one point

in each of the newly obtained squares and make sure no two points of all the points placed so far are on thesame horizontal or vertical line.

· · ·We continue this procedure infinitely and denote by A the collection of all the points placed according

to the above procedure. Then ∂A = [0, 1]× [0, 1] and A contains at most one point on each horizontal andeach vertical line.

I 1-18.

Proof. Clearly A ⊂ [0, 1]. For any x ∈ [0, 1]−A and any interval (a, b) with x ∈ (a, b), (a, b) must contain arational point of [0, 1]. So (a, b) ∩A = ∅ and (a, b) ∩Ac = ∅. This implies [0, 1]−A ⊂ ∂A. Since A is open,a boundary point of A cannot be in A. This implies ∂A ⊂ [0, 1]−A. Combined, we get [0, 1]−A = ∂A.

2

1.3 Functions and Continuity

2 Differentiation2.1 Basic DefinitionsI 2-4.

Proof. (a) If x = 0, then h(t) ≡ 0; if x = 0, h(t) = txg( x|x| ) by the property g(0, 1) = g(1, 0) = 0 andg(−x) = −g(x). Since h is a linear function, it is differentiable.

(b) f(x1, 0) = f(0, x2) ≡ 0 by the property g(0, 1) = g(1, 0) = 0 and g(−x) = −g(x). If f is differentiable,Df(0, 0) would have to be (0, 0). This implies |x| ·g( x|x| ) = f(x) = o(|x|) for x = 0. So g has to be identically0.

I 2-5.

Proof. Let g(x1, x2) = x1|x2| with (x1, x2) on unit circle. Then for (x, y) = 0,

g

((x, y)

|(x, y)|

)=

x√x2 + y2

· |y|√x2 + y2

=x|y|

x2 + y2.

Thereforef(x, y) = |(x, y)|g

((x, y)

|(x, y)|

).

I 2-6.

Proof. f(x, 0) = f(0, y) ≡ 0. So ∂f(0,0)∂x = ∂f(0,0)

∂y = 0. Assume f is differentiable at (0, 0), then

f(∆x,∆y)− f(0, 0) =∂f(0, 0)

∂x∆x+

∂f(0, 0)

∂y∆y + o(ρ) = o(ρ),

where ρ =√(∆x)2 + (∆y)2. This implies

√|∆x∆y| = o(

√(∆x)2 + (∆y)2). Let ∆y = ∆x, we get |∆x| =

o(√2|∆x|), contradiction.

I 2-8.

Proof. Note for any h ∈ R1 and λ = (λ1, λ2) ∈ R2, we have

maxi=1,2

{|f i(a+ h)− f i(a)− λih|} ≤ |f(a+ h)− f(a)− λh| ≤√2 maxi=1,2

{|f i(a+ h)− f i(a)− λih|}.

2.2 Basic TheoremsI 2-12.

Proof. (a)

f(h, k) = f(h1, · · · , hn, k1, · · · , km)

=n∑i=1

f(0, · · · , hi, · · · , 0, k1, · · · , km)

=n∑i=1

m∑j=1

f(0, · · · , hi, · · · , 0, 0, · · · , kj , · · · , 0)

=n∑i=1

m∑j=1

hikjf(0, · · · , 1, · · · , 0, 0, · · · , 1, · · · , 0).

3

So there existsM > 0, so that |f(h, k)| ≤M∑i,j |hikj | ≤M |(h, k)|2. This implies lim(h,k)→0 |f(h, k)|/|(h, k)| =

0.(b) f(a+ h, b+ k) = f(a, b+ k) + f(h, b+ k) = f(a, b) + f(a, k) + f(h, b) + f(h, k). So

lim|(h,k)|→0

|f(a+ h, b+ k)− f(a, b)− f(a, k)− f(h, b)||(h, k)|

= lim|(h,k)|→0

|f(h, k)||(h, k)|

= 0.

This implies Df(a, b)(x, y) = f(a, y) + f(x, b).

I 2-14.

Proof. We note

f(a1 + h1, a2 + h2, · · · , ak + hk)

= f(a1, · · · , ak) +k∑i=1

f(a1, · · · , ai−1, hi, ai+1, · · · , ak)

+k∑j=2

∑(x1, · · · , xk) consists of j h’s and k − j a’s

f(x1, · · · , xk).

Then use (a).

I 2-15.

Proof. (a) Note det is multi-linear, so we can use Problem 2.14(b).(b) Define xi(t) = (ai1(t), ai2(t), · · · , ain(t)). Then x′i(t) = (a′i1(t), · · · , a′in(t)) by Theorem 2.3(3) and f

can be seen as the composition g ◦ x with x(t) = (x1(t), · · · , xn(t))T ∈ Rn × · · · × Rn and g(x) = det(x).

Theorem 2-2 (chain rule) implies f ′(t) = Dg(x(t)) ◦ Dx(t). Since Dg(x)(y) =∑ni=1 det

x1· · ·y· · ·xn

and Dx =

x′1(t)· · ·x′n(t)

, we have

f ′(t) = Dg(x(t))(Dx(t)) =n∑i=1

det

x1(t)· · ·

Dx(t)i· · ·xn(t)

=n∑i=1

det

a11(t) · · · a1n(t)· · · · · · · · ·a′i1(t) · · · a′in(t)· · · · · · · · ·an1(t) · · · ann(t)

.

(c) Let A(t) =

a11(t) · · · a1n(t)· · · · · · · · ·an1(t) · · · ann(t)

, s(t) =

s1(t)· · ·sn(t)

and b(t) =

b1(t)· · ·bn(t)

. Then A(t)s(t) = b(t).

So s(t) = A−1(t)b(t) is differentiable. Moreover, A′(t)s(t) + A(t)s′(t) = b′(t). So s′(t) = A−1(t)(b′(t) −A′(t)A−1(t)b(t)) = A−1(t)b′(t)−A−1(t)A′(t)A−1(t)b(t).

2.3 Partial DerivativesI 2-23.

Proof. (b) f(x) = x21{x>0,y>0} − x21{x>0,y<0}.

4

2.4 DerivativesI 2-29.

Proof. (c)

Dxf(a) = limt→0

f(a+ tx)− f(a)

t= limt→0

f(a+ tx)− f(a)−Df(a)(tx)

|tx|· |tx|t

+Df(a)(x).

Note |tx|t is bounded for fixed x, and limt→0 |f(a+ tx)−f(a)−Df(a)(tx)|/|tx| = 0. So Dxf(a) = Df(a)(x).

I 2-30.

Proof. Note no matter t > 0 or t < 0, f(tx)−f(0)t = |x|g( x|x| ).

2.5 Inverse FunctionsI 2-36.

Proof. By Theorem 2-11, for any x ∈ A, there exists an open set Vx containing x and an open set Wx

containing f(x) such that f : Vx → Wx has a continuous inverse f−1 : Wx → Vx which is differentiable.Then f(A) = ∪x∈Af(Vx) = ∪x∈AWx must be open. For any open subset B of A, f(B) = (f−1)−1(B). Sincef−1 is continuous and B is open, (f−1)−1(B) must also be open.

Remark: For a proof without using Theorem 2-11, see [2] Theorem 8.2.

I 2-37.

Proof. We only prove part (a). Part (b) can be similarly proved. Assume f is 1-1, then the rank of Dfcannot be 1 in an open set. Indeed, if, for example, D1f(x, y) = 0 for all (x, y) in some open set A, considerg : A → R2 defined by g(x, y) = (f(x, y), y). Then detDg = 0 in A. By Inverse Function Theorem, g is 1-1in an open subset B of A and g(B) is open. So we can take two distinct points w1 and w2 in g(B) withthe same first coordinate. Suppose w1 = (f(x1, y1), y1) and w2 = (f(x2, y2), y2), then we must have y1 = y2and f(x1, y1) = f(x2, y2). This is contradictory with f being 1-1. So our claim must be true. (It is also astraightforward corollary of the rectification theorem, Theorem 2-13).

Consequently, for any given (x, y) ∈ R2 and any neighborhood of (x, y), there is at least one point (x′, y′)such that D1f(x

′, y′) = 0. By the continuity of D1f , we conclude D1f(x, y) = 0. Similarly, we can proveD2f(x, y) = 0. Combined, we have Df(x, y) = 0 for any (x, y) ∈ R2. This implies f is a constant (Problem2-22). So f cannot be 1-1 and we get contradiction again. Therefore, f cannot be 1-1.

Remark: The hint is basically Theorem 2.13 in the next section.

I 2-38.

Proof. (a) Assume not, then there exists x1, x2 ∈ R, such that x1 = x2 and f(x1) = f(x2). WLOG, assumex1 < x2, since f is differentiable everywhere, f is continuous. So there exists y, z ∈ [x1, x2] such thatf(y) = maxa∈[x1,x2] f(a) and f(z) = mina∈[x1,x2] f(a). Since f ′(a) = 0 for all a ∈ R, at least one of y, z isnot equal to x1 or x2. WLOG, assume y ∈ {x1, x2}. Then y ∈ (x1, x2). By Fermat’s theorem, f ′(y) = 0.Contradiction.

(b) f ′(x, y) =

(ex cos y −ex sin yex sin y ex cos y

). So detf ′(x, y) = e2x = 0. But clearly, f(x, y + 2nπ) = f(x, y)

(n = ±1,±2, · · · ). So f is not 1-1.

I 2-39.

Proof. f ′(x) ={

12 , if x = 012 + 2x sin 1

x − cos 1x , if x = 0

. Note as x → 0, 2x sin 1x → 0 and cos 1

x oscillates between

-1 and 1. So in any neighborhood of 0, f ′(x) becomes 0 infinitely many times. In particular, in anyneighborhood of 0, we can find a point x, so that f ′ has different signs on the two sides of x. This implies fis not 1-1 in a neighborhood of x.

5

2.6 Implicit FunctionsI 2-40.

Proof. Define f : R1 ×Rn → Rn as f(t, s1, · · · , sn) = (∑nj=1 aj1(t)sj − b1(t), · · · ,

∑nj=1 ajnsj − bn(t)). Then

(D1+jfi(t, s1, · · · , sn))1≤i,j≤n =

a11(t) a21(t) · · · an1(t)· · · · · · · · · · · ·an1(t) an2(t) · · · ann(t)

is non-singular. By Implicit Function Theorem, there is an open interval I containing t and an open setA containing (s1, · · · , sn) such that for each t ∈ I there is a unique s(t) ∈ A with f(t, s(t)) = 0 and s(t)differentiable.

3 Integration3.1 Basic Definitions3.2 Measure Zero and Content ZeroI 3-9.

Proof. (b) The set of integers.

I 3-10.

Proof. (a) By definition, ∀ε > 0, there is a finite cover {U1, · · · , Un} of A by closed rectangles such that∑ni=1 v(Ui) < ε. Clearly C ⊂ ∪ni=1Ui. So ∂C = C \ C0 ⊂ ∪ni=1Ui and hence the content of ∂C is 0.(b) C = [0, 1] ∩Q, then ∂C = [0, 1].

I 3-11.

Proof. Suppose ∂A has measure 0, then by the fact ∂A = [0, 1] − A and∑∞i=1(bi − ai) < 1, we conclude

[0, 1] = ∂A∪A has a measure less than 1. But [0, 1] is compact, so [0, 1] has a content less than 1, contradictorywith Theorem 3-5.

3.3 Integrable FunctionsI 3-14.

Proof. Use Theorem 3-8.

I 3-15.

Proof. By definition of content zero, C had content 0 implies C has content 0. So ∂C ⊂ C has content zeroand C is Jordan-measurable. By the definition of content zero, C is bounded, hence is contained by someclosed rectangle A. Finally, by Problem 3-8, C cannot contain any closed rectangle. So for any partition ofA, if S is a closed rectangle contained by A, minx∈S 1C(x) = 0. So

∫A1C(x)dx = L(1C , P ) = 0.

I 3-16.

Proof. As in Problem 3-10(b), C = [0, 1] ∩Q works since ∂C = [0, 1].

I 3-17.

Proof. Since C has measure 0, C cannot contain any closed rectangle by Theorem 3-6 and Problem 3-8. Sofor any sub-rectangle S of A, 1C obtains 0 on S. This implies L(1C , P ) = 0 for any partition P of A. Bydefinition of integration,

∫A1C = 0.

6

I 3-18.

Proof. It suffices to show Bn = {x : f(x) > 1/n} has content 0. Indeed, for any ε > 0, we can find a partitionP of A, such that

ε > U(f, P ) =∑S∈P

MS(f) ≥∑

S∈P,S∩Bn =∅

MS(f) >∑

S∈P,S∩Bn =∅

1

n· v(S).

So the collection of sets in P that intersect with Bn is a finite cover of Bn and has total content at most nε.Since ε is arbitrary, we conclude Bn has content 0.

I 3-19.

Proof. Let A = {x : f(x) = 1U (x)}. By Problem 1-8, ∂U = [0, 1] \ U . So ∀x ∈ ∂U \ A, f(x) = 1U (x) = 0.Fix x, by definition of boundary and the fact that U is open, any neighborhood of x must contain an intervalI ⊂ U . Since A has measure 0, I \ A = ∅. Therefore, any neighborhood of x contains points of U \ A, onwhich f takes the value 1. So f is discontinuous at x. That is, the set of discontinuous points of f contains∂U \A. Problem 3-11 shows ∂U does not have measure 0. So ∂U \A does not have measure 0. By Theorem3-8, f is not integrable on [0, 1].

I 3-20.

Proof. Use Problem 3-12 and Theorem 3-8.

I 3-21.

Proof. First of all, ∂C is bounded and closed, hence compact. By Theorem 3-6, A has measure 0 if and onlyif A has content 0. By Theorem 3-9, it suffices to show: ∂C has content 0 if and only if for every ε > 0,there is a partition P of A such that

∑S∈S1

v(S)−∑S∈S2

v(S) < ε, where S1 consists of all sub-rectanglesintersection C and S2 all sub-rectangles contained in C.

For necessity, we note ∂C has content 0, which implies ∀ε > 0, we can find finitely many closed rectangles{U1, · · · , Un} such that v(U1) + · · ·+ v(Un) < ε. We can extend {U1, · · · , Un} to a partition P of A, then

∑S∈S1

v(S)−∑S∈S2

v(S) ≤n∑i=1

v(Ui) < ε.

For sufficiency, it is clear that the condition implies ∂C has content 0.

I 3-22.

Proof. First note A is necessarily bounded by the definition of Jordan-measurable. Then use Problem 3-21:∫A−C

1 =

∫A

1−∫C

1 ≤∑S∈S1

v(S)−∑S∈S2

v(S) < ε.

7

3.4 Fubini’s TheoremI 3-23.

Proof. The key trick consists of two parts: for compact sets, content 0 is equivalent to measure 0; Problem3-18.

First, we can assume without loss of generality that C is closed. Indeed, denote by C the closureof C. Then by definition of content 0, for any ε > 0, there exists a finite cover {U1, · · · , Un} of C byclosed rectangles such that

∑ni=1 v(Ui) < ε. Since Ui’s are closed, they also consist of a cover of C. In

conclusion, C has content 0. Moreover, A′ = {x ∈ A : {y ∈ B : (x, y) ∈ C} is not of content 0} is containedby A′′ = {x ∈ A : {y ∈ B : (x, y) ∈ C} is not of content 0 }. So for our problem, it suffices to consider Cand prove A′′ has measure 0. Note C ⊂ A×B is bounded, so if C is closed, then C is compact.

By Problem 3-15, C is Jordan-measurable and∫A×B 1C = 0. By Fubini’s Theorem, for L(x) = L

∫B1C

and U(x) = U∫B1C , we have

∫A×B 1C =

∫AU =

∫AL. Again by Problem 3-15, 1A′U = U . Since C is

compact and coordinate projection is continuous, for any x ∈ A, {y : (x, y) ∈ C} is compact (Theorem 1-9).Therefore, on the set A′, U > 0 by Problem 3-18 and Theorem 3-6. On the other hand, 0 =

∫A×B 1C =∫

AU =

∫A1A′U . By Problem 3-18 and the fact U > 0 on A′, we conclude A′ has measure 0.

I 3-24.

Proof. C is the union of countably many segments: {1} × [0, 1], {12} × [0, 12 ], {

13} × [0, 13 ], {

23} × [0, 13 ], · · · .

So the rectangle [0, 1] × [0, 1n ] will cover all of these segments, except finitely many. These finitely many

segments can be covered by finitely many rectangles whose total volume is as small as we want. Therefore,C has content 0. Meanwhile A′ = Q ∩ [0, 1]. So A′ has measure 0 but is not of content 0.

I 3-25.

Proof. Since the set is compact, by Theorem 3.6, we can either prove the claim for content or prove the claimfor measure. When n = 1, Theorem 3-5 shows [a1, b1]× · · · × [an, bn] is not a set of measure 0. Suppose forn ≤ N , [a1, b1]× · · · × [an, bn] is not of measure 0, then∫

[a1,b1]×···×[aN+1,bN+1]

1 =

∫[aN+1,bN+1]

∫[a1,b1]×···×[aN ,bN ]

1 > 0

by Problem 3-18 and assumption. Mathematical induction concludes the claim is true for any n ≥ 1.

I 3-26.

Proof. The boundary of Af consists of four parts: Γ1 = {a}× [0, f(a)], Γ2 = {b}× [0, f(b)], Γ3 = [a, b]×{0},and Γ4 = {(x, y) : a ≤ x ≤ b, f(x−) ∧ f(x+) ≤ y ≤ f(x−) ∨ f(x+)}. It suffices to show Γ4 has measure0. Indeed, by the integrability of f , A = {x ∈ [a, b] : f is discontinuous at x} has measure 0. So ∀ε > 0,there exists open cover (Un)n≥1 of A, so that

∑∞n=1 v(Un) <

ε4M , where M = supa≤x≤b |f(x)|. Then

(Un × (−2M, 2M))n≥1 is an open cover of {(x, y) : x ∈ A, f(x−) ∧ f(x+) ≤ y ≤ f(x−) ∨ f(x+)} and∑∞n=1 v(Un×(−2M, 2M)) =

∑∞n=1 v(Un) ·4M < ε. The set B = [a, b]−∪∞

n=1Un is a compact set that consistof continuous points of f . By uniform continuity of f on B, we can find an open cover of {(x, f(x)) : x ∈ B}whose total volume is as small as we want. Combined, Γ4 ⊂ {(x, f(x)) : x ∈ ∪∞

n=1Un} ∪ {(x, f(x)) : x ∈ B}can be covered by open covers of arbitrarily small volume. So Γ4 has measure 0. This shows Af is Jordan-measurable. Its area is

∫ baf can be easily proved by Fubini’s Theorem.

I 3-27.

Proof.∫ ba

∫ yaf(x, y)dxdy =

∫ ba

∫ ba1{x≤y}f(x, y)dxdy. 1{x≤y} is clearly integrable. So 1{x≤y}f(x, y) is inte-

grable and Fubini’s Theorem applies:∫ ba

∫ ba1{x≤y}f(x, y)dxdy =

∫ ba

∫ ba1{x≤y}f(x, y)dydx =

∫ ba

∫ bxf(x, y)dydx.

I 3-28.

8

Proof. Assume D1,2f(a) − D2,1f(a) > 0, then by continuity, there is a rectangle A = [a1, b1] × [a2, b2]containing a such that D1,2f −D2,1f > 0 on A. By Fubini’s Theorem and Problem 3-18,

0 <

∫A

D1,2f −D2,1f =

∫ b1

a1

[D1f(x, b2)−D1f(x, a2)]−∫ b2

a2

[D2f(b1, y)−D1f(a1, y)] = 0.

Contradiction.

I 3-29.

Proof. Let A be a Jordan-measurable set in the yz-plane. For simplicity and for sake of illustration, weassume A is in the half plane {(y, z) : y ≥ 0}. Let Γ be the set of R3 obtained by revolving A about thez-axis. Then (x, y, z) ∈ Γ if and only if (

√x2 + y2, z) ∈ A. So∫

R3

χΓ(x, y, z)dxdydz =

∫R3

χA(√x2 + y2, z)dxdydz.

Using the change of variable {x = ρ cos θy = ρ sin θ,

we have∫R3

χΓ(x, y, z)dxdydz =

∫(ρ,θ,z)∈[0,∞)×[0,2π)×(−∞,∞)

χA(ρ, z)ρdρdθdz = 2π

∫ ∞

−∞

∫ ∞

0

χA(ρ, z)ρdρdz.

I 3-32.

Proof. To prove F ′(y) =∫ baD2f(x, y)dx, we only need to show for any c ≤ y1 ≤ y2 ≤ d, F (y2) − F (y1) =∫ y2

y1

∫ baD2f(x, y)dxdy. Since D2f is continuous, it is integrable. So Fubini’s Theorem implies∫ y2

y1

∫ b

a

D2f(x, y)dxdy =

∫ b

a

∫ y2

y1

D2f(x, y)dydx =

∫ b

a

[f(x, y2)− f(x, y1)]dx = F (y2)− F (y1).

In particular, our proof shows Leibnitz’s rule holds as far as D2f(x, y) is integrable on [a, b]× [c, d].

I 3-34.

Proof. By Leibnitz’s rule: D1f(x, y) = g1(x, 0) +∫ y0D1g2(x, t)dt = g1(x, 0) +

∫ y0D2g1(x, t)dt = g1(x, y).

I 3-35.

Proof. The key difficulty is to show any linear transformation is the composition of linear transformations ofthe type considered in (a). Recall the three linear transformations correspond to three basic manipulationsof matrices that reduce an invertible matrix to an identity matrix. So through the linear transformations ofthe type considered in (a), any identity matrix can be transformed into any invertible matrix.

9

3.5 Partitions of UnityI 3-37.

Proof. (a) Since f ≥ 0,∫ 1−εε

f is monotone increasing as ε → 0. So limε→0

∫ 1−εε

f exists if and onlyif limε→0

∫ 1−εε

f is bounded. Let Φ be a partition of unity, subordinate to an admissible cover O of (0, 1).Because [ε, 1−ε] is a compact set, only finitely many ϕ ∈ Φ are not 0 on [ε, 1−ε]. So

∫ 1−εε

f =∫ 1−εε

∑ϕ∈Φ ϕ ·

f =∑ϕ∈Φ

∫ 1−εε

ϕ · f ≤∑ϕ∈Φ

∫(0,1)

ϕ · f , and limε→0

∫ 1−εε

f ≤∑ϕ∈Φ

∫(0,1)

ϕ · f . On the other hand, for anyfinitely many ϕ in Φ, say ϕ1, ϕ2, · · · , ϕn, there exists ε > 0 such that all of ϕi’s (i = 1, · · · , n) are 0 outside[ε, 1− ε]. So

n∑i=1

∫(0,1)

ϕ · f =n∑i=1

∫ 1−ε

ε

ϕi · f =

∫ 1−ε

ε

n∑i=1

ϕi · f ≤∫ 1−ε

ε

f ≤ limε→0

∫ 1−ε

ε

f.

Let n→ ∞, we get∑ϕ∈Φ

∫ 1

0ϕ · f ≤ limε→0

∫ 1−εε

f . So by definition,

∫(0,1)

f exists ⇐⇒∑ϕ∈Φ

∫(0,1)

ϕ · f <∞ ⇐⇒ limε→0

∫ 1−ε

ε

f <∞ ⇐⇒ limε→0

∫ 1−εε

f exists.

(b) This problem is about the distinction between absolute convergence and conditional convergence ofthe series

∑∞n=1

(−1)n

n . Look up any standard textbook in analysis.

I 3-38.

Proof. We leave out the details and only describe the main idea. For each n, we can find a number an ∈(n, n + 1), such that

∫An∩(−∞,an]

f = 12(−1)n

n and∫An∩[an,∞)

f = (−1)n

2n . For a given small number ε > 0,the partition of unity in (n, n+ 1) can be chosen in two ways: for Φ, there are ϕ1 ∈ Φ and ϕ2 ∈ Φ such thatϕ1 ≡ 1 on (an−1, an − ε) and ϕ2 ≡ 1 on (an + ε, an+1), and they overlap a little bit around an; for Ψ, thereare ψ1 ∈ Ψ and ψ2 ∈ Ψ such that ψ1 ≡ 1 on (n, n+2− ε) and ψ2 ≡ 1 on (n+2+ ε, n+4), and they overlapa little bit around n+ 2. Then

∑ϕ∈Φ

∫ϕ · f ≈

∑∞n=−∞

12 [

(−1)n

n + (−1)n+1

n+1 ] = 12

∑∞n=−∞

(−1)n

n(n+1) is absolutelyconvergent, and

∑ψ∈Ψ

∫ψ ·f ≈

∑∞n=−∞[ (−1)n

n + (−1)n+1

n+1 ] =∑∞n=−∞

(−1)n

n(n+1) is absolutely convergent. Clearlythey converge to different values.

3.6 Change of VariableI 3-39.

Proof. Since g is continuous, the set B = {x ∈ A : detg′(x) = 0} is a closed set. Apply Theorem 3-13 toopen set A′ = A \ B, we have

∫g(A′)

f =∫A′(f ◦ g)|detg′| =

∫A(f ◦ g)|detg′|. By Sard’s Theorem, g(B) has

measure 0. So∫g(A)

f =∫g(A′)∪g(B)

f =∫g(A′)

f . So we still have Theorem 3-13 without the assumptiondetg′(x) = 0.

Remark: By Theorem 2-11 (Inver Function Theorem), if we stick to the condition detg′(x) = 0 in Theorem3-13, then g is an open mapping. So g(A) is open and talking about integration on g(A) is meaningful bythe definition of integrability in the extended sense. But without detg′(x) = 0, we cannot guarantee g is stillan open mapping and it may not be meaningful, rigorously speaking, to talk about integration on g(A) – wehave to check ∂g(A) has measure 0.

However, if B is bounded, it is compact and hence g(B) is compact. So ∂g(B) ⊂ g(B) has measure 0,and it is meaningful to talk about integration on ∂g(B).

10

4 Integration on Chains4.1 Algebraic PreliminariesI 4-1.

Proof. (a) By Theorem 4-4(3) and induction, when i1, · · · , ik are distinct,

φi1 ∧ · · · ∧ φik(ei1 , · · · , eik) = k!Alt(φi1 ⊗ · · · ⊗ φik)(ei1 , · · · , eik)

= k!1

k!

∑σ∈Sk

φi1 ⊗ · · · ⊗ φik(eσ(i1), · · · , eσ(ik)) · sgnσ

=∑σ∈Sk

φi1(eσ(i1)) · · ·φik(eσ(ik)) · sgnσ

= 1.

If the factor (k+l)!k!l! did not appear in the definition of ∧, the right hand side would be 1

k! .(b) Suppose vl =

∑nj=1 aljej (1 ≤ l ≤ k), then

φi1 ∧ · · · ∧ φik(v1, · · · , vk) = φi1 ∧ · · · ∧ φik(n∑j=1

a1jej , · · · ,n∑j=1

akjej)

=n∑j=1

φi1 ∧ · · · ∧ φik(a1jej ,n∑j=1

a2jej , · · · ,n∑j=1

akjej)

=k∑j=1

φi1 ∧ · · · ∧ φik(a1ijeij ,n∑j=1

a2jej , · · · ,n∑j=1

akjej)

= φi1 ∧ · · · ∧ φik(k∑j=1

a1ijeij ,n∑j=1

a2jej , · · · ,n∑j=1

akjej)

= · · ·

= φi1 ∧ · · · ∧ φik(k∑j=1

a1ijeij ,k∑j=1

a2ijeij , · · · ,k∑j=1

akijeij )

= det(alij )1≤l≤k,1≤j≤kφi1 ∧ · · · ∧ φik(ei1 , · · · , eik)= det(alij )1≤l≤k,1≤j≤k.

Remark: Combined with Theorem 4-6, this tells us how to calculate the result of a differential form actingon vectors.

I 4-2.

Proof. Λn(V ) has dimension 1. So f∗ as a linear mapping from Λn(V ) to Λn(V ) must be a multiplicationby some constant c. Let e1, · · · , en be a basis of V and φ1, · · · , φn the dual basis. Then

f∗(φ1 ∧ · · · ∧ φn)(e1, · · · , en) = φ1 ∧ · · · ∧ φn(f(e1), · · · , f(en)) = det(f) · φ1 ∧ · · · ∧ φn(e1, · · · , en)

by Theorem 4-6. By multi-linearity and alternating property, f∗(φ1 ∧ φn)(x1, · · · , xn) = det(f)φ1 ∧ · · · ∧φn(x1, · · · , xn) is true for any x1, · · · , xn ∈ V n.

I 4-3.

Proof. Let v1, · · · , vn be an orthonormal basis and wi =∑nj=1 aijvj . By Theorem 4-6, ω(w1, · · · , wn) =

det(aij)ω(v1, · · · , vn) = det(aij). Meanwhile, gij = T (wi, wj) = T (∑nk=1 aikvk,

∑nk=1 ajkvk) =

∑nk=1 aikajk.

If we denote the matrix (gij) by G and (aij) by A, we have G = AAT . So detG = (detA)2, i.e.

|ω(w1, · · · , wn)| =√

det(gij).

11

I 4-4.

Proof. We first show f(e1), · · · , f(en) is an orthonormal basis of V . Indeed, T (f(ei), f(ej)) = f∗T (ei, ej) =⟨ei, ej⟩ = δij . From this, we can easily show (f(ei))

ni=1 is linearly independent. Since dimV = n, (f(ei))ni=1 is

an orthonormal basis. Since [f(e1), · · · , f(en)] = µ, f∗ω(e1, · · · , en) = ω(f(e1), · · · , f(en)) = ω(e1, · · · , en) =1. By the multi-linearity and alternating property, f∗ω = det.

I 4-5.

Proof. Suppose ci(0) =∑nj=1 aij(t)c

j(t) (0 ≤ t ≤ 1). If we denote the matrix (aij(t)) by A(t), we have(c1(0), · · · , cn(0))T = A(t)(c1(t), · · · , cn(t))T . So det[(c1(0), · · · , cn(0))T ] = detA(t)det[(c1(t), · · · , cn(t))T ].Since A(t) is a continuous function of t, so is detA(t). Because A(t) is non-singular for any t ∈ [0, 1], detA(t)does not change sign. This implies [c1(0), · · · , cn(0)] = [c1(t), · · · , cn(t)], ∀t ∈ [0, 1].

I 4-6.

Proof. (b) Denote v1×· · ·×vn−1 by z. We first show v1, · · · , vn−1, z are linearly independent. Indeed, assume

not, then ⟨z, z⟩ = det

v1v2· · ·vn−1

z

= 0. So z = 0. On the other hand, by the linear independence of v1, · · · , vn−1,

there exists vn ∈ Rn such that v1, · · · , vn are linearly independent. As a result, ⟨vn, z⟩ = det

v1· · ·vn

= 0.

Contradiction. So v1, · · · , vn−1, z is a basis of Rn. Moreover, for any ω ∈ Λn(Rn), ω(v1, · · · , vn−1, z) =

det

v1· · ·vn−1

z

ω(e1, · · · , en) = ⟨z, z⟩ω(e1, · · · , en). So [v1, · · · , vn−1, z] is the usual orientation.

I 4-7.

Proof. For any non-zero ω ∈ Λn(V ), we can always find a basis v1, · · · , vn so that ω(v1, · · · , vn) = 1.Define a bilinear functional T on V × V by designating T (vi, vj) = δij . Then T can be extended toV × V and is an inner product. Under T , v1, · · · , vn is an orthonormal basis. Suppose v′1, · · · , v′n isanother basis of V , which is orthonormal under T and [v′1, · · · , v′n] = [v1, · · · , vn]. Then ω(v′1, · · · , v′n) =det(aij)ω(v1, · · · , vn) = det(aij), where A = (aij) is the representation matrix of v′1, · · · , v′n under v1, · · · ,vn. Since δij = T (v′i, v

′j) = T (

∑nk=1 aikvk,

∑nk=1 ajkvk) =

∑nk=1 aikajk, A = (aij) is an orthonormal matrix.

Hence |detA| = 1, which implies detA = 1.

I 4-8.

Proof. v1×· · ·× vn−1 is the unique element z ∈ V , such that for any w ∈ V , T (w, z) = ω(v1, · · · , vn−1, w) =

det

v1· · ·vn−1

w

ω(e1, · · · , en) = det

v1· · ·vn−1

w

.

I 4-9.

12

Proof. (a) We only prove e1 × e1 = 0 and e1 × e3 = −e2. Suppose e1 × e1 = z, then for any w ∈ V ,

⟨w, z⟩ = det

e1e1w

= 0. So z = 0. Suppose e1× e3 = a1e1+a2e2+a3e3, then for any w = b1e1+ b2e2+ b3e3,

⟨w, z⟩ = a1b1 + a2b2 + a3b3 = det

e1e3

b1e1 + b2e2 + b3e3

= b2det

e1e3e2

= −b2. Since b1, b2, b3 are arbitrary,

a1 = a3 = 0 and a2 = −1. That is e1 × e3 = −e2.(b) Suppose z = (z1, z2, z3), then

⟨z, v × w⟩ = det

v1 v2 v3

w1 w2 w3

z1 z2 z3

= z1(v2w3 − v3w2) + z2(w1v3 − v1w3) + z3(v1w2 − w1v2).

Since z is arbitrary, we conclude v × w = (v2w3 − v3w2, w1v3 − v1w3, v1w2 − w1v2).(c) Note sin2 θ = 1 − cos2 θ = 1 − ⟨v,w⟩2

|v||w|2 . So |v|2|w|2 sin2 θ = |v|2|w|2 − ⟨v, w⟩2 = |v × w|2, by part (b).⟨v × w, v⟩ = ⟨v × w,w⟩ = 0 is clear by definition of ”×”.

(d) ⟨v, w × z⟩ = ⟨w, z × v⟩ = ⟨z, v × w⟩ = det

vwz

= det

zvw

= det

wzv

. For rest of this part of

problem, use (b).(e) Use (c).

I 4.10.

Proof. Let V = span{w1, wn−1} and define φ ∈ Λn−1(V ) by φ(x1, · · · , xn−1) = det

x1· · ·xn−1

w1×···×wn−1

|w1×···×wn−1|

,

∀x1, · · · , xn−1 ∈ V . By definition of w1 × w2 × · · · × wn−1, φ(w1, · · · , wn−1) = |w1 × · · · × wn−1|. Ac-cording to Problem 4-5, it suffices to show φ is the volume element of V determined by ⟨, ⟩ and someorientation µ. Indeed, w1 × · · · × wn−1 is perpendicular to V , so for any orthonormal basis v1, · · · , vn−1 ofV , v1, · · · , vn−1, w1×···×wn−1

|w1×···×wn−1| is an orthonormal basis of Rn. This implies

φ(v1, vn−1) = det

v1· · ·vn−1

w1×···×wn−1

|w1×···×wn−1|

= ±1.

We can choose an orthonormal basis v∗1 , · · · , v∗n−1 in such a way that φ(v∗1 , · · · , v∗n−1) = 1. Denote[v∗1 , · · · , v∗n−1] by µ, then by Theorem 4-6, φ is the volume element determined by ⟨, ⟩ and µ.

I 4-11.

Proof. The property T (x, f(y)) = T (f(x), y) (∀x, y ∈ V ) holds if and only if T (vi, f(vj)) = T (f(vi), vj)(1 ≤ i, j ≤ n). Since f(vj) =

∑nk=1 ajkvk and f(vi) =

∑nk=1 aikvk, T (vi, f(vj)) =

∑nk=1 ajkT (vi, vk) = aji

and T (f(vi), vj) =∑nk=1 aikT (vk, vj) = aij . So we have aji = aij .

4.2 Fields and FormsI 4-13.

Proof. (a) For any vp ∈ Rnp ,

(g ◦ f)∗(vp) = (D(g ◦ f)(p)(v))g◦f(p) = (Dg(f(p)) ◦Df(p)(v))g◦f(p) = g∗[(Df(p)(v))f(p)] = g∗ ◦ f∗(vp).

13

For any ω ∈ Λk(Rpg◦f(p)),

(g ◦ f)∗(ω) = ω((g ◦ f)∗) = ω(g∗ ◦ f∗) = g∗ω(f∗) = f∗ ◦ g∗ω.

So (g ◦ f)∗ = f∗ ◦ g∗.(b) Apply Theorem 4-10(2) with k = l = 0.

I 4-14.

Proof. Let p = c(t), then f∗(v) = Df(p)(v) = (∑ni=1Dif

′(p)vi, · · · ,∑ni=1Dif

m(p)vi). Meanwhile, thetangent vector to f ◦ c at t is

((f ′ ◦ c)′(t), · · · , (fm ◦ c)′(t)) = (n∑i=1

Dif′(p)(ci)′(t), · · · ,

n∑i=1

Difm(p)(ci)′(t))

= (n∑i=1

Dif′(p)vi, · · · ,

n∑i=1

Difm(p)vi).

So the tangent vector to f ◦ c at t is f∗(v).

I 4-18.

Proof. Let g(t) = p+ tv, then

Dvf(p) = limt→0

f(p+ tv)− f(p)

t

= limt→0

f ◦ g(t)− f ◦ g(0)t

= Df(g(0))Dg(0)

= (D1f(p), · · · , Dnf(p))

v1· · ·vn

=

n∑i=1

Dif(p)vi

= ⟨v, w⟩.

After normalizing v to unit vector and by Cauchy-Schwartz inequality, Dvf(p)| ≤ |w| = D w|w|f(p). This

shows the gradient of f at p is the direction in which f is changing fastest at p.

I 4-20.

Proof. We only describe the intuition. We define a continuous transformation that pastes together thesegments AB and CD, where A = (0, 0), B = (1, 0), C = (1, 1), and D = (0, 1). The resulting image is aring. This solves the case n = 1.

For general n ≥ 2, divide the unit cube vertically into strips of equal size, and identify all strips with onewhile getting the orientation of edges “right”.

I 4-21.

Proof.∫c◦p ω =

∫ kI(c◦p)∗ω =

∫Ikp∗ ◦ c∗ω. Suppose c∗ω = hdx1∧· · ·∧dxn. So by change of variable formula

and Theorem 4-9,∫Ikp∗ ◦ c∗ω =

∫Ikh ◦ p detp′ =

∫Ikh ◦ p |detp′| =

∫Ikh =

∫Ikc∗ω =

∫c

ω.

14

4.3 Geometric PreliminariesI 4-24.

Proof. We fix a line L that goes through 0 and partition [0, 1] so that each c([ti−1, ti]) (1 ≤ i ≤ m) iscontained on one side of L. Let p(t) = {(x, y) : x2 + y2 = 1}∩Oc(t) where O denotes the origin (0, 0). Thenc([ti−1, ti]) and p([ti−1, ti]) are contained in the same side of L. Define a singular 2-cube ci ⊂ R2 − 0 as

ci(t, s) = tc(tis+ ti−1(t− s)) + (1− t)p(tis+ ti−1(1− s)).

Then

∂ci = [tc(ti−1) + (1− t)p(ti−1)] + c(tis+ ti−1(1− s))− [tc(ti) + (1− t)p(ti)]− p(tis+ ti−1(1− s)).

Define a singular 2-cube c2 ⊂ R2 − 0 by

c2(t, s) = ci(t,s− ti−1

ti − ti−1), if ti−1 ≤ s < ti.

Then it’s easy to see ∂c2 = c− c1,n for some integer n.Remark: Basically, the idea is to use homotopy to construct singular 2-cube in R2 − 0 and paste them

together through boundaries. After cancelation, ∂c2 becomes c− c1,n.

4.4 The Fundamental Theorem of CalculusI 4-26.

Proof. ∫cR,n

dθ =

∫[0,1]

c∗R,n

(−y

x2 + y2dx+

x

x2 + y2dy

)=

∫[0,1]

−R sin 2πnt

R2(R cos 2πnt)′dt+ R cos 2πnt

R2(R sin 2πnt)′dt

=

∫[0,1]

(sin 2πnt)22πn+ (cos 2πnt)22πn

= 2πn.

By Stokes’ Theorem, for any 2-chain c in R2 − 0,∫∂cdθ =

∫cd(dθ) = 0. This shows cR,n = ∂c.

I 4-27.

Proof. By c− c1 = ∂c2 and Stokes’ Theorem,∫cdθ −

∫c1,n

dθ =∫∂c2

dθ = 0. So∫cdθ =

∫c1,n

dθ = 2πn. Thisshows n is unique (=

∫cdθ/2π).

I 4-31.

Proof. Suppose for any ω = 0, we can find a chain c(ω) such that∫cω = 0. Then if d2ω = 0, we can find

a chain c such that 0 =∫cd2ω =

∫∂cdω =

∫∂2c

ω =∫0ω = 0. Contradiction. To construct c(ω), suppose

ω = fdx1 ∧ · · · dxn, then f ≡ 0. So we can find a point x0 such that f(x0) = 0. Let c(ω) be a sufficientlysmall cube centered at x0. Then

∫c(ω)

ω = 0.

I 4-32.

15

Proof. (a) c(s, t) = (1−t)c1(s)+tc2(s), then c(s, 0) = c1(s), c(s, 1) = c2(s), c(0, t) = c1(0), and c(1, t) = c1(1).Let c3 = c1(t) and c4 = c1(0), then ∂c = c1 − c2 + c3 − c4. This implies∫

∂c

ω =

∫c1

ω −∫c2

ω +

∫c3

ω −∫c4

ω.

If we suppose ω = u(x, y)dx + v(x, y)dy, then c∗3(ω) = u ◦ c3(c′3(t))′dt + v ◦ c3(c23(t))′dt = 0. Similarly,c∗4(ω) = 0. So

∫∂cω =

∫c1ω −

∫c2ω. There are two ways to show

∫∂cω = 0. First, if ω is well-defined

in c, then Stokes’ Theorem gives∫∂cω =

∫cdω = 0, provided ω is closed. Second, if ω is not well-defined

everywhere in c, but is exact, say ω = dω1, then Stokes’ Theorem gives∫∂c

ω =

∫∂c

dω1 =

∫∂2c

ω1 = 0.

Problem 4-21 gives the counter example.(b) Suppose ω = u(x, y)dx + v(x, y)dy. Choose a point (x0, y0) in the domain of ω, and define ω1 =∫ x

x0u(ξ, y0)dξ +

∫ yy0v(x, η)dη. This is the integration of ω along first [x0, x] and then [y0, y]. So dω1 =

u(x, y0)dx + v(x, y)dy +∫ yy0

∂∂xv(x, η)dη · dx, which implies ∂w1

∂y = v(x, y). By condition, we can also writeω1 as ω1 =

∫ yy0v(x0, η)dη +

∫ xx0u(ξ, y)dξ. This is the integration of ω along first [y0, y] and then [x0, x]. So

dω1 can also be written as dω1 = v(x0, y)dy + u(x, y)dx +∫ xx0

∂u∂y (ξ, y)dξ · dy, which implies ∂w1

∂x = u(x, y).Combined, we conclude dω1 = ω.

I 4-33.

Proof. (a) We only show f(z) = z is not analytic.

f(z)− f(z0)

z − z0=z − z0z − z0

=(x− x0)− i(y − y0)

(x− x0) + i(y − y0).

If z → z0 along the line x = x0, the limit is −1; if z → z0 along the line y = y0, the limit is 1. Solimz→z0

f(z)−f(z0)z−z0 does not exist.

(c) Let z = x + yi, then T (z) =

(a bc d

)(xy

)=

(ax+ bycx+ dy

). If z0 = x0 + y0i is a complex number

and T (z) = z0z = (x0 + y0i)(x + yi) = (x0x − y0y) + (x0y + y0x)i, we must have ax + by = x0x − y0y,cx + dy = x0y + y0x. Since x and y are arbitrary, this implies a = x0, b = −y0, c = y0 and d = x0, orequivalently, a = d, b = −c. By part(b) Cauchy-Riemann equations:

Df(z0) =

(Dxu(z0) Dyu(z0)Dxv(z0) Dyv(z0)

)=

(Dxu(z0) Dyu(z0)−Dyu(z0) Dxu(z0)

).

Df(z0) is a multiplication by the complex number Dxu(z0)− iDyu(z0).(d)

d(fdz) = d[(u+ iv)(dx+ idy)]

= d[udx− vdy + i(vdx+ udy)]

=∂u

∂ydy ∧ dx− ∂v

∂xdx ∧ dy + i[

∂v

∂ydy ∧ dx+

∂u

∂xdx ∧ dy]

= −[∂u

∂y+∂v

∂x]dx ∧ dy + i(

∂u

∂x− ∂v

∂y)dx ∧ dy.

So d(fdz) = 0 if and only if ∂u∂x = ∂v

∂y , ∂u∂y = − ∂v

∂x .(e) By part(d) and Stokes’ Theorem,

∫cfdz =

∫∂c′

fdz =∫∂c′

d(fdz) = 0.

16

(f)

gdz =1

x+ yi(dx+ idy) =

x− yi

x2 + y2(dx+ idy) =

xdx+ ydy

x2 + y2+xdy − ydx

x2 + y2i =

1

2d log(x2 + y2) + idθ.

Let h = 12 log(x2 + y2) : C− 0 → R. Then gdz = idθ + dh. So by Problem 4-26,∫

cR,n

dz

z=

∫cR,n

idθ + dh = 2πni+

∫[0,1]

c∗R,n(dh).

Since c∗R,n(dh) = dc∗R,n(h) = d 12 logR2 = 0. So

∫cR,n

dzz = 2πni.

(g) By Problem 4-23, Stokes’ Theorem and part(d), for some singular 2-cube c, cR1,n − cR2,n = ∂c and∫cR1,n−cR2,n

f(z)

zdz =

∫∂c

f(z)

zdz =

∫c

d

(f(z)

zdz

)= 0.

According to part(f), ∀ε > 0, when R is sufficiently small∣∣∣∣∣∫cR,n

f(z)

zdz − 2πinf(0)

∣∣∣∣∣ =∣∣∣∣∣∫cR,n

(f(z)

z− f(0)

z

)dz

∣∣∣∣∣ ≤ ε

∣∣∣∣∣∫cR,n

dz

z

∣∣∣∣∣ = ε2πn.

So limR→0

∫cR,n

f(z)z dz = 2πnif(0). By the first part of (g),

∫cR,n

f(z)z dz ≡ 2πnf(0)i. By Problem 4-24

(note we can further require c2 ⊂ C − 0),∫c−c1,n

f(z)z dz =

∫∂c2

f(z)z dz =

∫c2d(f(z)z dz

)= 0. So

∫cf(z)z dz =∫

c1,n

f(z)z dz = 2πnif(0), i.e. nf(0) = 1

2πi

∫cf(z)z dz = 0.

5 Integration on Manifolds5.1 ManifoldsI 5-1.

Proof. We note that ∂M in this problem is interpreted as “boundary of a manifold”, not “boundary of aset”.

For any x ∈ ∂M , there is an open set U containing x, an open set V ⊂ Rn, and a diffeomorphismh : U → V such that

h(U ∩ ∂M) = h(U ∩M ∩ ∂M) = V ∩ (Hk × {0}) ∩ {yk = 0} = {y ∈ V : yk = yk+1 = · · · = yn = 0}.

By definition, ∂M is a (k − 1)-dimensional manifold. Similarly, for any x ∈ M − ∂M , there is an open setU containing x, an open set V ⊂ Rn, and a diffeomorphism h : U → V such that

h(U∩(M−∂M)) = h(U∩M∩(M−∂M)) = V ∩(Hk×{0})∩{yk > 0} = {y ∈ V : yk > 0, yk+1 = · · · = yn = 0}.

Define V ′ = V ∩ {yk > 0} and U ′ = h−1(V ′). Then h remains a diffeomorphism from U ′ to V ′ andh(U ′ ∩ (M − ∂M)) = {y ∈ V ′ : yk+1 = · · · = yn = 0}. This shows M − ∂M is a k-dimensional manifold.

I 5-2.

Proof. See, for example, [2] §23, Example 3.

I 5-3. (a)

17

Proof. We first clarify that “boundary A” means set boundary. We show any point in boundary A belongsto one of the two types of points as illustrated by the problem’s counter example.

Indeed, for any point x in boundary A, by the definition of boundary A being an (n − 1)-dimensionalmanifold, we can find an open set U of Rn containing x, and a diffeomorphism h : U → V (V := h(U)) suchthat h(U ∩ boundary A) = {y ∈ V : yn = 0}. Without loss of generality, we assume U is connected. ThenV is also a connected open set. Since V − V ∩ {yn = 0} is the disjoint union of two connected open sets,U − boundary A = h−1(V − V ∩ {yn = 0}) is also the disjoint union of two connected open sets, say U1 andU2.

Since x is a (set) boundary point of A, any neighborhood that contains x must contains points of A.So at least one of U1 and U2 must contain points of A. Because A is open, after proper shrinking, we canassume that the Ui that contains points of A is entirely contained by A, i = 1, 2.

Then there are two cases to analyze. Case one, both U1 and U2 contain points of A. This impliesx ∈ U ⊂ [U1 ∪U2 ∪boundary A] ⊂ N . That is, x is an interior point of N and hence satisfies condition (M).Case two, only one of Ui’s contains points of A. Without loss of generality, we assume U1 contains pointsof A but U2 does not. Then x does not satisfy condition (M) but satisfies condition (M’). Combining thesetwo cases, we can conclude any point in boundary A satisfies either condition (M) or condition (M’).

It is clear that any point in A satisfies condition (M). So, we can conclude N is an n-dimensionalmanifold-with-boundary.

I 5-4.

Proof. For any x ∈ M , there is an open set U containing x, an open set V ⊂ Rn, and a diffeomorphismh : U → V such that

h(U ∩M) = V ∩ ({Rk × {0}}) = {y ∈ V : yk+1 = · · · = yn = 0}.

Let p : V → Rn−k be the projection onto the last (n− k) coordinates. Define g = p ◦ h. Then g : U → Rn−kis differentiable and g−1(0) = h−1 ◦ p−1(0) = h−1({y ∈ V : yk+1 = · · · = yn = 0}) = U ∩ M . SinceDg = Dp ◦Dh, rank of Dg(y) = rank of Dp(y) = n− k when g(y) = 0.

Remark: It’s a partial converse because g is only defined locally.

I 5-5.

Proof. Let X be a k-dimensional subspace of Rn. Choose an orthonormal basis a1, · · · , ak and extend it tobe an orthonormal basis a1, · · · , ak, ak+1, · · · , an of Rn. Define h to be the orthogonal transformation thatmaps X to {x ∈ Rn : xk+1 = · · · = xn = 0}. Then condition (M) is satisfied by h.

I 5-6.

Proof. Consider g : Rn × Rm → Rm defined by g(x, y) = y − f(x). Suppose f is differentiable, then gis differentiable and Dg(x, y) =

[Dxf(x) Im

]. So Dg has rank m. By Theorem 5-1, g−1(0) = {(x, y) :

y = f(x)} is an n-dimensional manifold. Conversely, if {(x, y) : y = f(x)} is a manifold, by Theorem 5-2f(x) is necessarily differentiable. Therefore the graph of f is an n-dimensional manifold if and only if f isdifferentiable.

I 5-8. (a)

Proof. Use the fact that in Rn, any open covering of a set has a countable sub-covering, i.e. the secondcountability axiom for a separable metric space.

(b)

Proof. For any x ∈ boundary M (set boundary), x ∈ M by the closedness of M . Clearly x cannot satisfycondition (M), so it must satisfy condition (M’). This implies x ∈ ∂M (manifold boundary). Conversely itis always true, by definition, that ∂M ⊂ boundary M . So boundary M = ∂M .

The counter example is already given in Problem 5-3(a).

(c)

18

Proof. By part (b), boundary M agrees with ∂M . By part (a) and Problem 5-1, ∂M has measure zero.Then we are done. (Compactness is used for boundedness of M , which is inherent in the definition ofJordan-measurability.)

5.2 Fields and Forms on ManifoldsI 5-9.

Proof. We suppose f is a coordinate system around x and f(a) = x. Let X be the collection of curves inM with c(0) = x. We need to show f∗(Rka) = span{c′(0) : c ∈ X}. Indeed, let X be the collection of curvesin Rk with c(0) = a, then f establishes a one-to-one correspondence between X and X by f : c→ f(c). ByProblem 4-14, the tangent vector to c = f(c) at 0 is f∗(c′(0)). So f∗(Rka) ⊃ span{c′(0) : c ∈ X}. For “c”,note f∗(ei) = c′i(0), where ci(t) = f(a+ tei).

I 5-10.

Proof. We first define an orientation on M . ∀x ∈ M , suppose f ∈ C is a coordinate system around x, suchthat f(a) = x for some a ∈ Rk. Define µx = [f∗((e1)a), · · · , f∗((ek)a)]. We need to check such an orientationµx is independent of the choice of f . Indeed, if g ∈ C is another coordinate patch around x with g(b) = x,by the fact det(f−1 ◦ g)′ > 0,

[(f−1 ◦ g)∗((e1)b), · · · , (f−1 ◦ g)∗((ek)b)] = [(e1)a, · · · , (ek)a].

Apply f∗ to both sides, we have

[g∗((e1)b), · · · , g∗((ek)b)] = [f∗((e1)a), · · · , f∗((ek)a)].

Second, by the very definition of µ, ∀f ∈ C, f is clearly orientation-preserving. Moreover, the requirementthat any element of C is orientation-preserving dictates that the orientation has to be defined in this way.

Finally, we show µ is consistent. Suppose h :W → Rn is a coordinate patch and a, b ∈W . Let x = h(a)and y = h(b). If [h∗((e1)a), · · · , h∗((ek)a)] = µh(a), there exists some f ∈ C, such that [h∗((e1)a), · · · , h∗((ek)a)] =µx = [f∗((e1)α), · · · , f∗((ek)α)] (f(α) = x = h(a)). This implies det(f−1 ◦ h)′(a) > 0 (see the argumenton page 118-119). Since both f and h are non-singular, det(f−1 ◦ h)′ does not change sign throughout itsdomain. In particular, if h(b) = y falls within the range of f , we must have det(f−1◦h)′(b) > 0, which implies[h∗((e1)b), · · · , h∗((ek)b)] = [f∗((e1)β), · · · , f∗((ek)β)] = µy (h(b) = f(β) = y). If h(b) = y is not within therange of f , we use a sequence of elements from C to “connect” x and y, provided M is connected.

I 5-16. Let g : A → Rp be as in Theorem 5-1. If f : Rn → R is differentiable and the maximum (orminimum) of f on g−1(0) occurs at a, show that there are λ1, · · · , λp ∈ R, such that

(1)Djf(a) =

p∑i=1

λiDjgi(a), j = 1, · · · , n.

Hint: This equation can be written df(a) =∑ni=1 λidg

i(a) and is obvious if g(x) = (xn−p+1, · · · , xn).The maximum of f on g−1(0) is sometimes called the maximum of f subject to the constraints gi = 0.

One can attempt to find a by solving the system of equations (1). In particular, if g : A→ R, we must solven+ 1 equations

Djf(a) = λDjg(a),

g(a) = 0,

in n+ 1 unknowns a1, · · · , an, λ, which is often very simple if we leave the equation g(a) = 0 for last. Thisis Lagrange’s method, and the useful but irrelevant λ is called a Lagrangian multiplier. The followingproblem gives a nice theoretical use for Lagrangian multipliers.

19

Proof. First, (1) can be re-written as Df = (λ1, · · · , λp)Dg, where Df and Dg are the Jacobian matricesof f and g, respectively. If g is the projection: g(x) = (xn−p+1, · · · , xn), then Dg(a) = (0p×(n−p), Ip×p) andg−1(0) = {x ∈ Rn : xn−p+1 = · · · = xn = 0}. So for any x ∈ g−1(0), f(x) = f(x1, · · · , xn−p, 0, · · · , 0). Whenf achieves its maximum (or minimum) on g−1(0) at a, we must have D1f(a) = · · · = Dn−pf(a) = 0. Defineλi = Dn−p+if(a) (1 ≤ i ≤ p), then

Df(a) = (0, · · · , 0, Dn−p+1f(a), · · · , Dnf(a))

= (0, · · · , 0, λ1, · · · , λp)= (λ1, · · · , λp)(0p×(n−p), Ip×p)

= (λ1, · · · , λp)Dg(a).

For general g, by Theorem 2-13, there is an open set U ⊂ Rn containing a and a differentiable functionh : U → Rn with differentiable inverse such that g ◦ h(x1, · · · , xn) = (xn−p+1, · · · , xn). Suppose h(x0) = a,then f ◦ h achieves its maximum (or minimum) on (g ◦ h)−1(0) at x0. By previous argument, there existsλ1, · · · , λp such that

D(f ◦ h)(x0) = (λ1, · · · , λp)D(g ◦ h)(x0).

By Theorem 2-2 (chain rule), D(f ◦h)(x0) = Df(a)Dh(x0) and D(g ◦h)(x0) = Dg(a)Dh(x0). Since Dh(x0)is invertible, we have Df(a) = (λ1, · · · , λp)Dg(a).

I 5-17. (a) Let T : Rn → Rn be self-adjoint with matrix A = (aij), so that aij = aji. If f(x) = ⟨Tx, x⟩ =∑aijx

ixj , show that Dkf(x) = 2∑nj=1 akjx

j . By considering the maximum of ⟨Tx, x⟩ on Sn−1 show thatthere is x ∈ Sn−1 and λ ∈ R with Tx = λx.

(b) If V = {y ∈ Rn : ⟨x, y⟩ = 0}, show that T (V ) ⊂ V and T : V → V is self-adjoint.(c) Show that T has a basis of eigenvectors.

(a)

Proof. Define g(x) = |x|2 − 1. Then g is a differentiable function with g−1(0) = Sn−1 = {x ∈ Rn : |x| = 1}.So g′(x) = 2x has rank 1 whenever g(x) = 0. Sn−1 is a compact set, so f(x) = ⟨Tx, x⟩ achieves its maximumand minimum on Sn−1, say at xmax and xmin, respectively. By Problem 5-16, there exists λ ∈ R, suchthat Df(xmin) = λDg(xmin) and Df(xmax) = λDg(xmax). Note Dkf(x) =

∑nj=1 akjx

j +∑ni=1 aikx

i =

2∑nj=1 akjx

j = 2(x1, · · · , xn)

ak1· · ·akn

, we can re-write the above equation as

(x10, · · · , xn0 )

a11 a21 · · · an1· · · · · · · · · · · ·a1n a2n · · · ann

= λx0,

where x0 = xmin or xmax. This shows xmin and xmax are eigenvectors of T on Sn−1. Since x0 ∈ {x : ||x|| = 1},λ = ⟨Ax0, x0⟩. This is the so-called Maximum Principle. See Principle of Applied Mathematics, by RobertKeener, page 17.

(b)

Proof. ∀y ∈ V , ∠x, Ty⟩ = ⟨Tx, y⟩ = λ⟨x, y⟩ = 0. So T (V ) ⊂ V . Since the corresponding matrix A of T issymmetric, T : V → V is self-adjoint.

(c)

Proof. Part(a) shows T has at least one eigenvector x1. Part(b) shows T remains a self-adjoint linearoperator on the (n − 1)-dimensional space {x1}⊥. So we can find a second eigenvector x2 ∈ {x1}⊥. ThenV ′ = {y ∈ Rn : ⟨x1, y⟩ = ⟨x2, y⟩ = 0} = {x1, x2}⊥ = {x1}⊥ ∩ {x2}⊥ again satisfies the properties enjoyedby V in part(b). So we can find a third eigenvector x3 ∈ V ′. Continue this procedure, we can find a basisconsisting of eigenvectors of T .

20

5.3 Stokes’ Theorem on Manifolds5.4 The Volume ElementI 5-23.

Proof. Suppose c∗(ds) = α(t)dt. By definition,

α(t) = α(t)dt(1)

= c∗(ds)(1)

= ds(c∗(1))

= ds(((c1)′, · · · , (cn)′))

=

√√√√ n∑i=1

[(ci)′]2ds

(((c1)′, · · · , (cn)′)√∑n

i=1[(ci)′]2

)

=

√√√√ n∑i=1

[(ci)′]2,

where the last “=” is by the definition of length element and the fact that c is orientation-preserving.

I 5-24.

Proof. dV (e1, · · · , en) = 1 = dx1 ∧ · · · ∧ dxn(e1, · · · , en).

I 5-25.

Proof. Suppose M is an oriented (n− 1)-dimensional manifold in Rn. ∀x ∈M and let n(x) be the outwardunit normal to M . We define ω ∈ Λn−1(Mx) by

ω(v1, · · · , vn−1) = det

n(x)v1

· · ·vn−1

, ∀v1, · · · , vn−1 ∈Mx.

So if v1, · · · , vn−1 is an orthonormal basis of Mx with [n(x), v1, · · · , vn−1] equal to the usual orientation ofRnx , ω(v1, · · · , vn−1) = 1. This shows dA = ω.

On the other hand, Mx is a subspace of Rnx . So Λn−1(Mx) is a subspace of Λn−1(Rnx). This implies dAcan be represented as dA = α1dx

2∧· · ·∧dxn+(−1)α2dx1∧dx3∧· · ·∧dxn+ · · ·+(−1)n+1αndx

1∧· · ·∧dxn−1.

So ∀v1, · · · , vn−1 ∈Mx, by Problem 4-1(b), dA(v1, · · · , vn−1) = det

αv1

· · ·vn−1

. So αi(−1)1+i = ni.

I 5-26.

Proof. (a) Define c : [a, b]× [0, 2π] → R3 by c(x, θ) = (x, f(x) cos θ, f(x) sin θ). So

Dc =

1 0f ′(x) cos θ −f(x) sin θf ′(x) sin θ f(x) cos θ

.

Using the notation in the text, we have E = 1 + (f ′(x))2, G = f2(x), F = 0. so the area elementdA =

√1 + (f ′(x))2f(x)dx ∧ dθ and

∫SdA =

∫ ba2πf(x)

√1 + (f ′(x))2dx.

21

(b) Let a = −1, b = 1, f(x) =√1− x2, then

Area(S2) =

∫ 1

−1

2π√1− x2

1√1− x2

dx = 4π.

I 5-27.

Proof. Denote by dV1 and dV2 the volume elements of M and T (M), respectively. The problem is reducedto proving dV1 = T ∗(dV2). Indeed, ∀x ∈ M and suppose (v1, · · · , vn) is an orthonormal basis of Mx with[v1, · · · , vn] = µx. Then (T (v1), T (v2), · · · , T (vn)) is an orthonormal basis of T (M)T (x) as well. To see[T (v1), T (v2), · · · , T (vn)] = T (µx), note if v = Aw, then T (v) = TAw = TAT−1(Tw). Combined, bydefinition of volume element, we have T ∗(dV2)(v1, · · · , vn) = dV2(T (v1), · · · , T (vn)) = 1. By the uniquenessof volume element on M , we conclude T ∗(dV2) = dV1.

5.5 The Classical Theorems

References[1] Peter Lax. Linear algebra. John Wiley & Sons, Inc., New York, 1997. 2

[2] J. Munkres. Analysis on manifolds, Westview Press, 1997. 5, 17

22