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Analysis on Manifolds
Solution of Exercise Problems
Yan Zeng
Version 0.1.1, last revised on 2014-03-25.
Abstract This is a solution manual of selected exercise problems from Analysis on manifolds , by James R. Munkres [1]. If you find any typos/errors, please email me at zypublic@hotmail.com.
Contents
1 Review of Linear Algebra 3
2 Matrix Inversion and Determinants 3
3 Review of Topology in Rn^ 4
4 Compact Subspaces and Connected Subspace of Rn^ 5
5 The Derivative 5
6 Continuously Differentiable Functions 5
7 The Chain Rule 6
8 The Inverse Function Theorem 6
9 The Implicit Function Theorem 6
10 The Integral over a Rectangle 6
11 Existence of the Integral 7
12 Evaluation of the Integral 7
13 The Integral over a Bounded Set 7
14 Rectifiable Sets 7
15 Improper Integrals 7
16 Partition of Unity 7
17 The Change of Variables Theorem 7
18 Diffeomorphisms in Rn^ 7
- 19 Proof of the Change of Variables Theorem
- 20 Applications of Change of Variables
- 21 The Volume of a Parallelepiped
- 22 The Volume of a Parametrized-Manifold
- 23 Manifolds in Rn
- 24 The Boundary of a Manifold
- 25 Integrating a Scalar Function over a Manifold
- 26 Multilinear Algebra
- 27 Alternating Tensors
- 28 The Wedge Product
- 29 Tangent Vectors and Differential Forms
- 30 The Differential Operator
- 31 Application to Vector and Scalar Fields
- 32 The Action of a Differentiable Map
- 33 Integrating Forms over Parametrized-Manifolds
- 34 Orientable Manifolds
- 35 Integrating Forms over Oriented Manifolds
- 36 A Geometric Interpretation of Forms and Integrals
- 37 The Generalized Stokes’ Theorem
- 38 Applications to Vector Analysis
- 39 The Poincaré Lemma
- 40 The deRham Groups of Punctured Euclidean Space
- 41 Differentiable Manifolds and Riemannian Manifolds
Proof. B =
b 11 b 12 b 13 b 21 b 22 b 23
. Then BA =
b 11 + b 12 2 b 11 b 12 + b 13 b 21 + b 22 2 b 21 b 12 + b 23
. So BA = I 2 if and only if
8
<
:
b 11 + b 12 = 1 b 21 + b 22 = 0 2 b 11 b 12 + b 13 = 0 2 b 21 b 22 + b 23 = 1:
Plug b 12 = b 11 1 and b 22 = b 21 into the las two equations, we have
{ 3 b 11 + b 13 = 1 3 b 21 + b 23 = 1:
So we can have the following two different left inverses for A: B 1 =
and B 2 =
(b)
Proof. By Theorem 2.2, A has no right inverse.
Proof. (a) By Theorem 1.5, n m and among the n row vectors of A, there are exactly m of them are linearly independent.[ By applying elementary row operations to A, we can reduce A to the echelon form Im 0
]
. So we can find a matrix D that is a product of elementary matrices such that DA =
[
Im 0
]
(b) If rankA = m, by part (a) there exists a matrix D that is a product of elementary matrices such that
DA =
[
Im 0
]
Let B = [Im; 0]D, then BA = Im, i.e. B is a left inverse of A. Conversely, if B is a left inverse of A, it is easy to see that A as a linear mapping from Rm^ to Rn^ is injective. This implies the column vectors of A are linearly independent, i.e. rankA = m. (c) A has a right inverse if and only if Atr^ has a left inverse. By part (b), this implies rankA = rankAtr^ = n.
Proof. Suppose (Dk)Kk=1 is a sequence of elementary matrices such that DK D 2 D 1 A = In. Note DK D 2 D 1 A = DK D 2 D 1 InA, we can conclude A ^1 = DK D 2 D 1 In.
Proof. A ^1 =
d b c a
1 d bc by Theorem 2.14.
3 Review of Topology in Rn
Proof. X = R, Y = (0; 1], and A = Y.
Proof. For any closed subset C of Y , f ^1 (C) = [f ^1 (C) \ A] [ [f ^1 (C) \ B]. Since f ^1 (C) \ A is a closed subset of A, there must be a closed subset D 1 of X such that f ^1 (C) \ A = D 1 \ A. Similarly, there is a closed subset D 2 of X such that f ^1 (C) \ B = D 2 \ B. So f ^1 (C) = [D 1 \ A] [ [D 2 \ B]. A and B are closed in X, so D 1 \ A, D 2 \ B and [D 1 \ A] [ [D 2 \ B] are all closed in X. This shows f is continuous.
Proof. (a) Take f (x) y 0 and let g be such that g(y 0 ) ̸= z 0 but g(y)! z 0 as y! y 0.
4 Compact Subspaces and Connected Subspace of Rn
Proof. (a) Let xn = (2n+ 2 ) ^1 and yn = (2n 2 ) ^1. Then as n! 1, jxn ynj! 0 but j sin (^) x^1 n sin (^) y^1 n j =
Proof. The boundedness of X is clear. Since for any i ̸= j, jjei ej jj = 1, the sequence (ei)^1 i=1 has no accumulation point. So X cannot be compact. Also, the fact jjei ej jj = 1 for i ̸= j shows each ei is an isolated point of X. Therefore X is closed. Combined, we conclude X is closed, bounded, and non- compact.
5 The Derivative
Proof. By definition, limt! 0 f^ (a+tu t) f^ (a)exists. Consequently, limt! 0 f^ (a+tu t) f^ (a)= limt! 0 f^ (a+tcu ct) f^ (a) exists and is equal to cf ′(a; u).
Proof. (a) f (u) = f (u 1 ; u 2 ) = (^) uu 21 u^2 1 +u^22
. So
f (tu) f (0) t
t
t^2 u 1 u 2 t^2 (u^21 + u^22 )
t
u 1 u 2 u^21 + u^22
In order for limt! 0 f^ (tu) t f^ (0)to exist, it is necessary and sufficient that u 1 u 2 = 0 and u^21 + u^22 ̸= 0. So for vectors (1; 0) and (0; 1), f ′(0; u) exists, and we have f ′(0; (1; 0)) = f ′(0; (0; 1)) = 0. (b) Yes, D 1 f (0) = D 2 f (0) = 0. (c) No, because f is not continuous at 0 : lim(x;y)! 0 ;y=kx f (x; y) = kx
2 x^2 +k^2 x^2 =^
k 1+k^2. For^ k^ ̸= 0, the limit is not equal to f (0). (d) See (c).
6 Continuously Differentiable Functions
Proof. We note jxyj √ x^2 + y^2
x^2 + y^2 √ x^2 + y^2
x^2 + y^2 :
So lim(x;y)! 0 pjxyj x^2 +y^2 = 0. This shows f (x; y) = jxyj is differentiable at 0 and the derivative is 0. However,
for any fixed y, f (x; y) is not a differentiable function of x at 0. So its partial derivative w.r.t. x does not exist in a neighborhood of 0 , which implies f is not of class C^1 in a neighborhood of 0.
11 Existence of the Integral
12 Evaluation of the Integral
13 The Integral over a Bounded Set
14 Rectifiable Sets
15 Improper Integrals
16 Partition of Unity
17 The Change of Variables Theorem
18 Diffeomorphisms in R n
19 Proof of the Change of Variables Theorem
20 Applications of Change of Variables
21 The Volume of a Parallelepiped
- (a)
Proof. Let v = (a; b; c), then XtrX = (I 3 ; vtr)
I 3
v
= I 3 +
a b c
A (^) (a; b; c) =
1 + a^2 ab ac ab 1 + b^2 bc ca cb 1 + c^2
A.
(b)
Proof. We use both methods:
V (X) = [det(Xtr^ X)]1/2^ = [(1 + a^2 )(1 + b^2 + c^2 ) ab ab + ca ( ac)]1/2^ = (1 + a^2 + b^2 + c^2 )1/
and
V (X) =
(^4) det^2 I 3 + det^2
a b c
A (^) + det^2
a b c
A (^) + det^2
a b c
A
1/ = (1 + c^2 + a^2 + b^2 )1/2:
Proof. Let X = (x 1 ; ; xi; ; xk) and Y = (x 1 ; ; xi; ; xk). Then V (Y ) = [
[I] det
2 Y
I ]1/2^ =
[
[I]
(^2) det^2 XI ]1/2 (^) = jj[∑ [I] det
2 X
I ]^
(^12) = jjV (X).
Proof. Suppose P is determined by x 1 , , xk. Then V (h(P)) = V (x 1 ; ; xk) = jjV (x 1 ; x 2 ; ; xk) = = jjkV (x 1 ; x 2 ; ; xk) = jjkV (P).
- (a)
Proof. Straightforward.
(b)
Proof.
jjajj^2 jjbjj^2 ⟨a; b⟩^2 = (
∑^3
i=
a^2 i )(
∑^3
j=
b^2 j ) (
∑^3
k=
akbk)^2
∑^3
i;j=
a^2 i b^2 j
∑^3
k=
a^2 kb^2 k 2(a 1 b 1 a 2 b 2 + a 1 b 1 a 3 b 3 + a 2 b 2 a 3 b 3 )
∑^3
i;j=1;i̸=j
a^2 i b^2 j 2(a 1 b 1 a 2 b 2 + a 1 b 1 a 3 b 3 + a 2 b 2 a 3 b 3 )
= (a 2 b 3 a 3 b 2 )^2 + (a 1 b 3 a 3 b 1 )^2 + (a 1 b 2 a 2 b 1 )^2
= det^2
a 2 b 2 a 3 b 3
a 1 b 1 a 3 b 3
a 1 b 1 a 2 b 2
- (a)
Proof. Suppose V 1 and V 2 both satisfy conditions (i)-(iv). Then by the Gram-Schmidt process, the uniqueness is reduced to V 1 (x 1 ; ; xk) = V 2 (x 1 ; ; xk), where x 1 , , xk are orthonormal.
(b)
Proof. Following the hint, we can assume without loss of generality that W = Rn^ and the inner product is the dot product on Rn. Let V (x 1 ; ; xk) be the volume function, then (i) and (ii) are implied by Theorem 21.4, (iii) is Problem 2, and (iv) is implied by Theorem 21.3: V (x 1 ; ; xk) = [det(XtrX)]1/2.
22 The Volume of a Parametrized-Manifold
Proof. By definition, v(Z ) =
A V^ (D^ ). Let^ x^ denote the general point of^ A; let^ y^ =^ (x)^ and^ z^ =^ h◦^ (x) = (y). By chain rule, D (x) = Dh(y) D (x). So [V (D (x))]^2 = det(D (x)trDh(y)trDh(y)D (x)) = [V (D (x))]^2 by Theorem 20.6. So v(Z ) =
A V^ (D^ ) =^
A V^ (D^ ) =^ v(Y^ ).
Proof. Let x denote the general point of A. Then
D (x) =
0 B B B B B B @
D 1 f (x) D 2 f (x) Dkf (x)
1 C C C C C C A
and by Theorem 21.4, V (D (x)) =
[
∑k i=1(Dif^ (x))
2
]1/
. So v(Y ) =
A
[
∑k i=1(Dif^ (x))
2
]1/
- (a)
23 Manifolds in Rn
Proof. In this case, we set U = R and V = M = f(x; x^2 ) : x 2 Rg. Then maps U onto V in a one-to-one fashion. Moreover, we have (1) is of class C^1. (2) ^1 ((x; x^2 )) = x is continuous, for (xn; x^2 n)! (x; x^2 ) as n! 1 implies xn! x as n! 1.
(3) D (x) =
[
2 x
]
has rank 1 for each x 2 U. So M is a 1-manifold in R^2 covered by the single coordinate patch.
Proof. We let U = H^1 and V = N = f(x; x^2 ) : x 2 H^1 g. Then maps U onto V in a one-to-one fashion. Moreover, we have (1) is of class C^1. (2) ^1 ((x; x^2 )) = x is continuous.
(3) D (x) =
[
2 x
]
has rank 1 for each x 2 H^1. So N is a 1-manifold in R^2 covered by the single coordinate patch.
- (a)
Proof. For any point p 2 S^1 with p ̸= (1; 0), we let U = (0; 2 ), V = S^1 (1; 0), and : U! V be defined by () = (cos ; sin ). Then maps U onto V continuously in a one-to-one fashion. Moreover, (1) is of class C^1. (2) ^1 is continuous, for (cos n; sin n)! (cos ; sin ) as n! 1 implies n! as n! 1.
(3) D () =
[
sin cos
]
has rank 1.
So is a coordinate patch. For p = (1; 0), we consider U = ( ; ), V = S^1 ( 1 ; 0), and : U! V be defined by () = (cos ; sin ). We can prove in a similar way that is a coordinate patch. Combined, we can conclude the unit circle S^1 is a 1-manifold in R^2.
(b)
Proof. We claim ^1 is not continuous. Indeed, for tn = 1 (^) n^1 , (tn)! (1; 0) on S^1 as n! 1, but (^1) ( (tn)) = tn! 1 ̸= (^1) ((1; 0)) = 0 as n! 1.
Proof. Let U = A and V = f(x; f (x)) : x 2 Ag. Define : U! V by (x) = (x; f (x)). Then maps U onto V in a one-to-one fashion. Moreover, (1) is of class Cr. (2) ^1 is continuous, for (xn; f (xn))! (x; f (x)) as n! 1 implies xn! x as n! 1.
(3) D (x) =
[
Ik Df (x)
]
has rank k.
So V is a k-manifold in Rk+1^ with a single coordinate patch.
Proof. For any x 2 M and y 2 N , there is a coordinate patch for x and a coordinate patch for y, respectively. Denote by U the domain of , which is open in Rk^ and by W the domain of , which is open in either Rl^ or Hl. Then U W is open in either Rk+l^ or Hk+l, depending on W is open in Rl^ or Hl. This is the essential reason why we need at least one manifold to have no boundary: if both M and N have boundaries, U W may not be open in Rk+l^ or Hk+l.
The rest of the proof is routine. We define a map f : U W! (U ) (W ) by f (x; y) = ( (x); (y)). Since (U ) is open in M and (W ) is open in N by the definition of coordinate patch, f (U W ) = (U ) (W ) is open in M N under the product topology. f is one-to-one and continuous, since and enjoy such properties. Moreover, (1) f is of class Cr, since and are of class Cr^. (2) f ^1 = ( ^1 ; ^1 ) is continuous since ^1 and ^1 are continuous.
(3) Df (x; y) =
[
D (x) 0 0 D (y)
]
clearly has rank k + l for each (x; y) 2 U W. Therefore, we conclude M N is a k + l manifold in Rm+n.
- (a)
Proof. We define 1 : [0; 1)! [0; 1) by 1 (x) = x and 2 : [0; 1)! (0; 1] by 2 (x) = x + 1. Then it’s easy to check 1 and 2 are both coordinate patches.
(b)
Proof. Intuitively I I cannot be a 2-manifold since it has “corners”. For a formal proof, assume I I is a 2-manifold of class Cr^ with r 1. By Theorem 24.3, @(I I), the boundary of I I, is a 1-manifold without boundary of class Cr. Assume is a coordinate patch of @(I I) whose image includes one of those corner points. Then D cannot exist at that corner point, contradiction. In conclusion, I I cannot be a 2-manifold of class Cr^ with r 1.
24 The Boundary of a Manifold
Proof. The equation for the solid torus N in cartesian coordinates is (b
x^2 + y^2 )^2 + z^2 a^2 , and the
equation for the torus T in cartesian coordinates is (b
x^2 + y^2 )^2 + z^2 = a^2. Define O = R and f : O! R
by f (x; y; z) = a^2 z^2 (b
x^2 + y^2 )^2. Then Df (x; y; z) =
2 x px^2 xb (^2) +y 2 2 y p^2 yb x^2 +y^2 2 z
5 has rank^1 at each point of
T. By Theorem 24.4, N is a 3-manifold and T = @N is a 2-manifold without boundary.
Proof. We first prove a regularization result.
Lemma 24.1. Let f : Rn+k^! Rn^ be of class Cr_. Assume_ Df has rank n at a point p , then there is an open set W Rn+k^ and a Cr -function G : W! Rn+k^ with Cr -inverse such that G(W ) is an open neighborhood of p and f ◦ G : W! Rn^ is the projection mapping to the first n coordinates.
Proof. We write any point x 2 Rn+k^ as (x 1 ; x 2 ) with x 1 2 Rn^ and x 2 2 Rk. We first assume Dx 1 f (p) has
rank n. Define F (x) = (f (x); x 2 ), then DF =
[
Dx 1 f Dx 2 f 0 Ik
]
. So detDF (p) = detDx 1 f (p) ̸= 0. By the
inverse function theorem, there is an open set U of Rn+k^ containing p such that F carries U in a one-to-one fashion onto an open set W of Rn+k^ and its inverse function G is of class Cr. Denote by : Rn+k^! Rn^ the projection (x) = x 1 , then f ◦ G(x) = ◦ F ◦ G(x) = (x) on W. In general, since Df (p) has rank n, there will be j 1 < < jn such that the matrix (^) @@(x(fj 11 ;;^ ;f;xnjn) (^) ) has rank
n at p. Here xj^ denotes the j-th coordinate of x. Define H : Rn+k^! Rn+k^ as the permutation that swaps the pairs (x^1 ; xj^1 ), (x^2 ; xj^2 ), , (xn; xjn^ ), i.e. H(x) = (xj^1 ; xj^2 ; ; xjn^ ; ) (pj^1 ; pj^2 ; ; pjn^ ; ) + p. Then
H(p) = p and D(f ◦H)(p) = Df (H(p))DH(p) = Df (p)DH(p). So Dx 1 (f ◦H)(p) = (^) @@(x(fj 11 ;;^ ;f;xnjn) (^) ) (p) and f ◦H is of the type considered previously. So using the notation of the previous paragraph, f ◦ (H ◦ G)(x) = (x) on W.
Proof. We note
Df (x) =
@(f 1 ; ; f 6 ) @(x 11 ; x 12 ; x 13 ; x 21 ; x 22 ; x 23 ; x 31 ; x 32 ; x 33 )
2 x 11 2 x 12 2 x 13 0 0 0 0 0 0 0 0 0 2 x 21 2 x 22 2 x 23 0 0 0 0 0 0 0 0 0 2 x 31 2 x 32 2 x 33 x 21 x 22 x 23 x 11 x 12 x 13 0 0 0 x 31 x 32 x 33 0 0 0 x 11 x 12 x 13 0 0 0 x 31 x 32 x 33 x 21 x 22 x 23
Since x 1 , x 2 , x 3 are pairwise orthogonal and are non-zero, we conclude x 1 , x 2 and x 3 are independent. From the structure of Df , the row space of Df (x) for x 2 O(3) has rank 6. By the theorem proved in Problem 2, O(3) is a 3-manifold without boundary in R^9. Finally, O(3) = fx : f (x) = 0g is clearly bounded and closed, hence compact.
Proof. The argument is similar to that of Problem 5, and the dimension = n^2 n n(n 2 1)= n(n 2 1).
25 Integrating a Scalar Function over a Manifold
Proof. To see (t; z) is a coordinate patch, we note that is one-to-one and onto S^2 (a) D, where D = f(x; y; z) : (
p a^2 z^2 ; 0 ; z); jzj ag is a closed set and has measure zero in S^2 (a) (note D is a parametrized 1-manifold, hence it has measure zero in R^2 ). On the set f(t; z) : 0 < t < 2 ; jzj < ag, is smooth and (^1) (x; y; z) = (t; z) is continuous on S (^2) (a) D. Finally, by the calculation done in the text, the rank of D
is 2 on f(t; z) : 0 < t < 2 ; jzj < ag.
(D )trD
[
(a^2 z^2 )1/2^ sin t (a^2 z^2 )1/2^ cos t 0 ( z cos t)/(a^2 z^2 )1/2^ ( z sin t)/(a^2 z^2 )1/2^1
]
(a^2 z^2 )1/2^ sin t ( z cos t)/(a^2 z^2 )1/ (a^2 z^2 )1/2^ cos t ( z sin t)/(a^2 z^2 )1/ 0 1
[
a^2 z^2 0 a
2 a^2 z^2
]
So V (D ) = a and v(S^2 (a)) =
f(t;z):0<t< 2 ;jzj<ag V^ (D^ ) = 4a
Proof. Let ( (^) j ) be a family of coordinate patches that covers M. Then (h ◦ (^) j ) is a family of coordinate patches that covers N. Suppose ϕ 1 , , ϕl is a partition of unity on M that is dominated by ( (^) j ), then
ϕ 1 ◦ h ^1 , , ϕl ◦ h ^1 is a partition of unity on N that is dominated by (h ◦ (^) j ). Then
∫
N
f dV =
∑^ l
i=
N
(ϕi ◦ h ^1 )f dV
∑^ l
i=
IntUi
(ϕi ◦ h ^1 ◦ h ◦ (^) i)(f ◦ h ◦ (^) i)V (D(h ◦ (^) i))
∑^ l
i=
IntUi
(ϕi ◦ (^) i)(f ◦ h ◦ (^) i)V (D (^) i)
∑^ l
i=
M
ϕi(f ◦ h)dV
M
f ◦ hdV:
In particular, by setting f 1 , we get v(N ) = v(M ).
Proof. Let L 0 = fx 2 Rn^ : xi > 0 g. Then M \ L 0 is a manifold, for if : U! V is a coordinate patch on M , : U \ ^1 (L 0 )! V \ L 0 is a coordinate patch on M \ L. Similarly, if we let L 1 = fx 2 Rn^ : xi < 0 g, M \ L 1 is a manifold. Theorem 25.4 implies
ci(M ) =
v(M )
M
dV =
v(M )
[∫
M \L 0
dV +
M \L 1
dV
]
Suppose ( (^) j ) is a family of coordinate patches on M \ L 0 and there is a partition of unity ϕ 1 ; ; ϕl on M \ L 0 that is dominated by ( (^) j ), then
∫
M \L 0
idV =
∑^ l
j=
M
(ϕj i)dV =
∑^ l
j=
IntUj
(ϕj ◦ (^) j )(i ◦ (^) j )V (D (^) j )
Define f : Rn^! Rn^ by f (x) = (x 1 ; ; xi; ; xn). It’s easy to see (f ◦ (^) j ) is a family of coordinate patches on M \ L 1 and ϕ 1 ◦ f , , ϕl ◦ f is a partition of unity on M \ L 1 that is dominated by (f ◦ (^) j ). Therefore
∫
M \L 1
idV =
∑^ l
j=
IntUj
(ϕj ◦f ◦f ◦ (^) j )(i ◦f ◦ (^) j )V (D(f ◦ (^) j )) =
∑^ l
j=
IntUj
(ϕj ◦ (^) j )(i ◦f ◦ (^) j )V (D(f ◦ (^) j ))
In order to show ci(M ) = 0, it suffices to show (i ◦ (^) j )V (D (^) j ) = (i ◦ f ◦ (^) j )V (D(f ◦ (^) j )). Indeed,
V 2 (D(f ◦ (^) j ))(x) = V 2 (Df ( (^) j (x))D (^) j (x)) = det(D (^) j (x)trDf ( (^) j (x))trDf ( (^) j (x))D (^) j (x)) = det(D (^) j (x)trD (^) j (x)) = V 2 (D )(x);
and i ◦ f = i. Combined, we conclude
M \L 1 idV^ =^ ^
M \L 0 idV^. Hence^ ci(M^ ) = 0.
- (a)
Proof. Let ( (^) i) be a family of coordinate patches on M and ϕ 1 , , ϕl a partition of unity on M dominated by ( (^) i). Let ( (^) j ) be a family of coordinate patches on N and 1 , , (^) k a partition of unity on N dominated
27 Alternating Tensors
Proof. Since h is not multilinear, h is not an alternating tensor. f = ϕ 1 ; 2 ϕ 2 ; 1 + ϕ 1 ; 1 is a tensor. The only permutation of f 1 ; 2 g are the identity mapping id and : (1) = 2; (2) = 1. So f is alternating if and only if f ^ (x; y) = f (x; y). Since f ^ (x; y) = f (y; x) = y 1 x 2 y 2 x 1 + y 1 x 1 ̸= f (x; y), we conclude f is not alternating. Similarly, g = ϕ 1 ; 3 ϕ 3 ; 2 is a tensor. And g^ = ϕ 2 ; 1 ϕ 2 ; 3 ̸= g. So g is not alternating.
Proof. Suppose I = (i 1 ; ; ik). If fi 1 ; ; ikg̸ = fj 1 ; ; jkg (set equality), then ϕI (aj 1 ; ; ajk ) = 0. If fi 1 ; ; ikg = fj 1 ; ; jkg, there must exist a permutation of f 1 ; 2 ; ; kg, such that I = (i 1 ; ; ik) = (j(1); ; j(k)). Then ϕI (aj 1 ; ; ajk ) = (sgn)(ϕI )^ (aj 1 ; ; ajk ) = (sgn)ϕI (ajσ(1) ; ; ajσ(k) ) = sgn. In summary, we have
ϕI (aj 1 ; ; ajk ) =
sgn if there is a permutation of f 1 ; 2 ; ; kg such that I = J = (j(1); ; j(k)) 0 otherwise.
Proof. For any v 1 ; ; vk 2 V and a permutation of f 1 ; ; kg.
(T f )^ (v 1 ; ; vk) = T f (v(1); ; v(k)) = f (T (v(1)); ; T (v(k))) = f ^ (T (v 1 ); ; T (vk)) = (sgn)f (T (v 1 ); ; T (vk)) = (sgn)T f (v 1 ; ; vk):
So (T f )^ = (sgn)T f , which implies T f 2 Ak(V ).
Proof. We follow the hint and prove ϕIσ = (ϕI )
1
. Indeed, suppose a 1 ; ; an is a basis of the underlying vector space V , then
(ϕI )
1 (aj 1 ; ; ajk ) = (ϕI )(ajσ (^1) (1) ; ; ajσ (^1) (k) ) =
0 if I ̸= (j (^1) (1); ; j (^1) (k)) 1 if I = (j (^1) (1); ; j (^1) (k))
0 if I ̸= (j◦ (^1) (1); ; j◦ (^1) (k)) = J 1 if I = (j◦ (^1) (1); ; j◦ (^1) (k)) = J = ϕIσ (aj 1 ; ; ajk ):
Thus, ϕI =
(sgn)(ϕI^ )
^1 (sgn
(^1) )(ϕI ) ^1 = ∑ ^1 (sgn)ϕIσ^ =^
(sgn)ϕIσ^.
28 The Wedge Product
- (a)
Proof. F = 2ϕ 2 ϕ 2 ϕ 1 +ϕ 1 ϕ 5 ϕ 4 , G = ϕ 1 ϕ 3 +ϕ 3 ϕ 1. So AF = 2ϕ 2 ^ϕ 2 ^ϕ 1 +ϕ 1 ^ϕ 5 ^ϕ 4 = ϕ 1 ^ϕ 4 ^ϕ 5 and AG = ϕ 1 ^ ϕ 3 ϕ 1 ^ ϕ 3 = 0, by Step 9 of the proof of Theorem 28.1.
(b)
Proof. (AF ) ^ h = ϕ 1 ^ ϕ 4 ^ ϕ 5 ^ (ϕ 1 2 ϕ 3 ) = 2ϕ 1 ^ ϕ 4 ^ ϕ 5 ^ ϕ 3 = 2ϕ 1 ^ ϕ 3 ^ ϕ 4 ^ ϕ 5.
(c)
Proof. (AF )(x; y; z) = ϕ 1 ^ ϕ 4 ^ ϕ 5 (x; y; z) = det
x 1 y 1 z 1 x 4 y 4 z 4 x 5 y 5 z 5
(^5) = x 1 y 4 z 5 + x 1 y 5 z 4 + x 4 y 1 z 5 x 4 y 5 z 1
x 5 y 1 z 4 + x 5 y 4 z 1.
Proof. Suppose G is a k-tensor, then AG(v 1 ; ; vk) =
(sgn)G (^) (v 1 ; ; vk) = ∑ (sgn)G(v^1 ;^ ^ ; vk) = [
(sgn)]G(v^1 ;^ ^ ; vk). Let^ e^ be an elementary permutation. Then^ e^ :^ ^!^ e^ ◦^ ^ is an isomorphism on the permutation group Sk of f 1 ; 2 ; ; kg. So Sk can be divided into two disjoint subsets U 1 and U 2 so that e∑ establishes a one-to-one correspondence between U 1 and U 2. By the fact sgne ◦ = sgn, we conclude
(sgn) = 0. This implies^ AG^ = 0.
Proof. We work by induction. For k = 2, (^) l 1!^1 l 2! A(f 1 f 2 ) = f 1 ^ f 2 by the definition of ^. Assume for k = n, the claim is true. Then for k = n + 1,
1 l 1! ln!ln+1!
A(f 1 fn fn+1) =
l 1! ln!
ln+1!
A((f 1 fn) fn+1) =
l 1! ln!
A(f 1 fn)^fn+
by Step 6 of the proof of Theorem 28.1. By induction, (^) l 1 !^1 ln! A(f 1 fn) = f 1 ^ ^ fn. So 1 l 1 !ln!ln+1! A(f^1 ^ fn^ fn+1) =^ f^1 ^ ^^ fn^ ^^ fn+1. By the principle of mathematical induction,
1 l 1! lk!
A(f 1 fk) = f 1 ^ ^ fk
for any k.
Proof. ϕi 1 ^ ϕik (x 1 ; ; xk) = A(ϕi 1 ϕik )(x 1 ; ; xk) =
(sgn)(ϕi 1 ^ ϕik )
(^) (x ∑^1 ;^ ^ ; xk) = (sgn)(ϕi^1 ^ ϕik )(x(1);^ ^ ; x(k)) =^
(sgn)xi 1 ;(1);^ ^ ; xik ;(k)^ =^ detXI^.
Proof. Suppose F is a k-tensor. Then
T (F ^ )(v 1 ; ; vk) = F ^ (T (v 1 ); ; T (vk)) = F (T (v(1)); ; T (v(k))) = T F (v(1); ; v(k)) = (T F )^ (v 1 ; ; vk):
- (a)
Proof. T ^ I (v 1 ; ; vk) = (^) I (T (v 1 ); ; T (vk)) = (^) I (B v 1 ; ; B vk). In particular, for J = (j 1 ; ; jk), c (^) J¯ =
[J] cJ^ J^ (e¯j 1 ;^ ^ ; e¯jk ) =^ T^ (^) I (e¯j 1 ;^ ^ ; e¯jk ) =^ I^ (B^ ^ e¯j 1 ;^ ^ ; B^ ^ e¯jk ) =^ I^ (^ ¯j 1 ;^ ^ ;^ ¯jk )^ where^ i^ is the i-th column of B. So c (^) J¯ = det[ ¯j 1 ; ; ¯jk ]I. Therefore, cJ is the determinant of the matrix consisting of the i 1 , , ik rows and the j 1 , , jk columns of B, where I = (i 1 ; ; ik) and J = (j 1 ; ; jk).
(b)
Proof. T f =
[I] dI^ T^
( I ) = ∑
[I] dI
[I] detBI;J^ J^ =^
[J](
[I] dI^ detBI;J^ )^ J^ where^ BI;J^ is the matrix consisting of the i 1 , , ik rows and the j 1 , , jk columns of B (I = (i 1 ; ; ik) and J = (j 1 ; ; jk)).
30 The Differential Operator
Proof. d! = xdx ^ dy zdy ^ dz. So d(d!) = dx ^ dx ^ dy dz ^ dy ^ dz = 0. Meanwhile,
d = 2 yzdz ^ dy + 2dx ^ dz = 2yzdy ^ dz + 2dx ^ dz
and ! ^ = ( xy^2 z^2 3 x)dx ^ dy + (2x^2 y + xyz)dx ^ dz + (6x y^2 z^3 )dy ^ dz:
So d(! ^ ) = ( 2 xy^2 z 2 x^2 xz + 6)dx ^ dy ^ dz; (d!) ^ = 2 x^2 dx ^ dy ^ dz xzdx ^ dy ^ dz;
and ! ^ d = 2xy^2 zdx ^ dy ^ dz 6 dx ^ dy ^ dz:
Therefore, (d!) ^ ! ^ d = ( 2 xy^2 z 2 x^2 xz + 6)dx ^ dy ^ dz = d(! ^ ).
Proof. In R^2 ,! = ydx xdy vanishes at x 0 = (0; 0), but d! = 2 dx ^ dy does not vanish at x 0. In general, suppose! is a k-form defined in an open set A of Rn, and it has the general form! =
[I] fI^ dxI^. If it vanishes at each x in a neighborhood of x 0 , we must have fI = 0 in a neighborhood of x 0 for each I. By continuity, we conclude fI 0 in a neighborhood of x 0 , including x 0. So d! =
[I] dfI^ ^^ dxI^ =^
[I](
i Dif dxi)^ ^^ dxI vanishes at x 0.
Proof. d! = d
x x^2 +y^2 dx
y x^2 +y^2 dy
= (^) (x (^22) +xyy (^2) ) 2 dx ^ dy + (^) (x (^2) +^2 xyy (^2) ) 2 dx ^ dy = 0. So! is closed. Define
= 12 log(x^2 + y^2 ), then d = !. So! is exact on A.
- (a)
Proof. d! = (x
(^2) +y (^2) )+2y 2 (x^2 +y^2 )^2 dy^ ^^ dx^ +^
x^2 +y^2 2 x^2 (x^2 +y^2 )^2 dx^ ^^ dy^ = 0. So^!^ is closed. (c)
Proof. We consider the following transformation from (0; 1 ) (0; 2 ) to B: { x = r cos t y = r sin t:
Then
det @(x; y) @(r; t)
= det
[
cos t r sin t sin t r cos t
]
= r ̸= 0:
By part (b) and the inverse function theorem (Theorem 8.2, the global version), we conclude ϕ is of class C^1.
(d)
Proof. Using the transformation given in part (c), we have dx = cos tdr r sin tdt and dy = sin tdr +r cos tdt. So! = [ r sin t(cos tdr r sin tdt) + r cos t(sin tdr + r cos tdt)]/r^2 = dt = dϕ.
(e)
Proof. We follow the hint. Suppose g is a closed 0 -form in B. Denote by a the point ( 1 ; 0) of R^2. For any x 2 B, let (t) : [0; 1]! B be the segment connecting a and x, with (0) = a and (1) = x. Then by mean-value theorem (Theorem 7.3), there exists t 0 2 (0; 1), such that g(a) g(x) = Dg(a+t 0 (x a))(a x). Since g is closed in B, Dg = 0 in B. This implies g(x) = g(a) for any x 2 B.
(f)
Proof. First, we note ϕ is not well-defined in all of A, so part (d) can not be used to prove! is exact in A. Assume! = df in A for some 0 -form f. Then d(f ϕ) = df dϕ =! ! = 0 in B. By part (e), f ϕ is a constant in B. Since limy# 0 ϕ(1; y) = 0 and limy" 0 ϕ(1; y) = 2, f (1; y) has different limits when y approaches 0 through positive and negative values. This is a contradiction since f is C^1 function defined everywhere in A.
Proof. d =
∑n i=1( 1) i (^1) Difidxi ^ dx 1 ^ dxci ^ ^ dxn = ∑n i=1 Difidx^1 ^ ^^ dxn. So^ d^ = 0^ if and only if
∑n i=1 Difi^ = 0. Since^ Difi(x) =^
jjxjj^2 mx^2 i jjxjjm+2^ ,^
∑n i=1 Difi(x) =^
n m jjxjjm^. So^ d^ = 0^ if and only if^ m^ =^ n.
Proof. By linearity, it suffices to prove the theorem for! = f dxI , where I = (i 1 ; ; ik 1 ) is a k- tuple from f 1 ; ; ng in ascending order. Indeed, in this case, h(x) = d(f dxI )(x)((x; v 1 ); ; (x; vk)) = (
∑n i=1 Dif^ (x)dxi^ ^^ dxI^ )((x;^ v^1 );^ ^ ;^ (x;^ vk)).^ Let^ X^ = [v^1 ^ vk].^ For each^ j^ 2 f^1 ;^ ^ ; kg, let^ Yj^ = [v 1 bvj vk]. Then by Theorem 2.15 and Problem 4 of §28,
detX(i; i 1 ; ; ik 1 ) =
∑^ k
j=
( 1)j ^1 vij detYj (i 1 ; ; ik 1 ):
Therefore
h(x) =
∑^ n
i=
Dif (x)detX(i; i 1 ; ; ik 1 )
∑^ n
i=
∑^ k
j=
Dif (x)( 1)j ^1 vij detYj (i 1 ; ; ik 1 )
∑^ k
j=
( 1)j ^1 Df (x) vj detYj (i 1 ; ; ik 1 ):
Meanwhile, gj (x) = !(x)((x; v 1 ); ; (̂x; vj ); ; (x; vk)) = f (x)detYj (i 1 ; ; ik 1 ). So
Dgj (x) = Df (x)detYj (i 1 ; ; ik 1 )
and consequently, h(x) =
∑k j=1( 1) j (^1) Dgj (x) vj. In particular, for k = 1, h(x) = Df (x) v, which is a
directional derivative.
31 Application to Vector and Scalar Fields
Proof. (Proof of Theorem 31.1) It is straightforward to check that (^) i and (^) j are isomorphisms. Moreover, d ◦ 0 (f ) = df =
∑n i=1 Dif dxi^ and^1 ◦^ grad(f^ ) =^1 ((x;^
∑n i=1 Dif^ (x)ei)) =^
∑n i=1 Dif^ (x)dxi. So^ d^ ◦^0 = 1 ◦^ grad.
