Chapter 1. Smooth Manifolds

Theorem 1. [Exercise 1.18] Let M be a topological manifold. Then any two smoothatlases for M determine the same smooth structure if and only if their union is a smoothatlas.

Proof. Suppose A1 and A2 are two smooth atlases for M that determine the samesmooth structure A. Then A1,A2 ⊆ A, so A1 ∪A2 must be a smooth atlas since everychart in A1 is compatible with every chart in A2. Conversely, if A1 ∪ A2 is a smoothatlas then the smooth structures determined by A1 and A2 both contain A1 ∪A2. Butthere is exactly one smooth structure containing A1 ∪A2, so A1 and A2 determine thesame smooth structure.

Theorem 2. [Exercise 1.44] Let M be a smooth n-manifold with boundary and let Ube an open subset of M .

(1) U is a topological n-manifold with boundary, and the atlas consisting of allsmooth charts (V, ϕ) for M such that V ⊆ U defines a smooth structure onU . With this topology and smooth structure, U is called an open submanifoldwith boundary.

(2) If U ⊆ IntM , then U is actually a smooth manifold (without boundary); in thiscase we call it an open submanifold of M .

(3) IntM is an open submanifold of M (without boundary).

Proof. Parts (1) and (2) are obvious. Part (3) follows from (2) and the fact that IntMis an open subset of M .

Theorem 3. [Problem 1-6] Let M be a nonempty topological manifold of dimensionn ≥ 1. If M has a smooth structure, then it has uncountably many distinct ones.

Proof. We use the fact that for any s > 0, Fs(x) = |x|s−1 x defines a homeomorphismfrom Bn to itself, which is a diffeomorphism if and only if s = 1. Let A be a smoothstructure for M and choose some coordinate map ϕ : U → Bn centered at some p ∈ U .Let A′ be the smooth atlas obtained by replacing every coordinate map ψ : V → R inA with ψ′ : V ∩ (M \ p) → R, except when ψ = ϕ. For any s > 0, let As be thesmooth atlas obtained from A′ by replacing ϕ with Fs ϕ. Since Fs is a diffeomorphismif and only if s = 1, the smooth structures determined by As and At are the same if andonly if s = t. This shows that there are uncountably many distinct smooth structureson M .

Theorem 4. [Problem 1-8] An angle function on a subset U ⊆ S1 is a continuousfunction θ : U → R such that eiθ(z) = z for all z ∈ U . There exists an angle function θon an open subset U ⊆ S1 if and only if U 6= S1. For any such angle function, (U, θ) isa smooth coordinate chart for S1 with its standard smooth structure.

1

2

Proof. The first part follows from the lifting criterion (Proposition A.78). By rotatingR2 appropriately, we may assume that N = (0, 1) /∈ U . Let σ : S1 \ N → R be thestereographic projection given by σ(x1, x2) = x1/(1− x2). We can compute

(σ θ−1)(α) =cosα

1− sinα,

which is a diffeomorphism on θ(U).

Theorem 5. [Problem 1-10] Let k and n be integers satisfying 0 < k < n, and letP,Q ⊆ Rn be the linear subspaces spanned by (e1, . . . , ek) and (ek+1, . . . , en), respec-tively, where ei is the ith standard basis vector for Rn. For any k-dimensional subspaceS ⊆ Rn that has trivial intersection with Q, the coordinate representation ϕ(S) con-structed in Example 1.36 is the unique (n− k)× k matrix B such that S is spanned by

the columns of the matrix

(IkB

), where Ik denotes the k × k identity matrix.

Proof. Let B = e1, . . . , ek. The matrix of ϕ(S) represents the linear map (πQ|S) (πP |S)−1. Since πP |S is an isomorphism, the vectors (πP |S)−1(B) form a basis for S.But then

(IdRn |S (πP |S)−1)(B) = ((πP |S + πQ|S) (πP |S)−1)(B)

= (IdP +πQ|S (πP |S)−1)(B)

is a basis for S. This is precisely the result desired.

Theorem 6. [Problem 1-12] Suppose M1, . . . ,Mk are smooth manifolds and N is asmooth manifold with boundary. Then M1 × · · · ×Mk × N is a smooth manifold withboundary, and ∂(M1 × · · · ×Mk ×N) = M1 × · · · ×Mk × ∂N .

Proof. Denote the dimensions ofM1, . . . ,Mk, N by n1, . . . , nk, d. Let p = (p1, . . . , pk, q) ∈M1 × · · · ×Mk ×N , let ϕi : Mi → Rni be coordinate maps around pi for i = 1, . . . , k,and let ψ : N → Hd be a coordinate map around q. Then ϕ1×· · ·×ϕk×ψ : M1×· · ·×Mk × N → Hn1+···+nk+d is a coordinate map, and (ϕ1 × · · · × ϕk × ψ)(p1, . . . , pk, q) ∈∂Hn1+···+nk+d if and only if ψ(q) ∈ ∂Hd, which is true if and only if q ∈ ∂N .

Chapter 2. Smooth Maps

Theorem 7. [Exercise 2.1] Let M be a smooth manifold with or without boundary.Then pointwise multiplication turns C∞(M) into a commutative ring and a commutativeand associative algebra over R.

Proof. The product of two smooth functions is also smooth.

3

Theorem 8. [Exercise 2.2] Let U be an open submanifold of Rn with its standardsmooth manifold structure. Then a function f : U → Rk is smooth in the sense ofsmooth manifolds if and only if it is smooth in the sense of ordinary calculus.

Proof. Obvious since the single chart IdRn covers Rn.

Theorem 9. [Exercise 2.3] Let M be a smooth manifold with or without boundary, andsuppose f : M → Rk is a smooth function. Then f ϕ−1 : ϕ(U) → Rk is smooth forevery smooth chart (U,ϕ) for M .

Proof. If p ∈ U then there is a smooth chart (V, ψ) for M such that f ψ−1 is smoothand p ∈ V . But on ϕ(U ∩ V ) we have

f ϕ−1 = f ψ−1 ψ ϕ−1,

which is smooth since ψ ϕ−1 is smooth. This shows that f ϕ−1 is smooth in aneighborhood of every point in ϕ(U), so f ϕ−1 is smooth.

Theorem 10. [Exercise 2.7(1)] Suppose M and N are smooth manifolds with or withoutboundary, and F : M → N is a map. Then F is smooth if and only if either of thefollowing conditions is satisfied:

(1) For every p ∈ M , there exist smooth charts (U,ϕ) containing p and (V, ψ)containing F (p) such that U ∩ F−1(V ) is open in M and the composite mapψ F ϕ−1 is smooth from ϕ (U ∩ F−1(V )) to ψ(V ).

(2) F is continuous and there exist smooth atlases (Uα, ϕα) and (Vβ, ψβ) for Mand N , respectively, such that for each α and β, ψβ F ϕ−1

α is a smooth mapfrom ϕα (Uα ∩ F−1(Vβ)) to ψβ(Vβ).

Proof. (1) is obvious, as is (2)⇒ (1). If (1) holds then F is continuous by Proposition2.4, and we can choose the smooth atlases to be the smooth structures for M and N .For every α and β the map ψβ F ϕ−1

α is smooth since for all p ∈ M we can findsmooth coordinate maps ϕ containing p and ψ containing F (p) such that

ψβ F ϕ−1α = ψβ ψ−1 ψ F ϕ ϕ−1 ϕ−1

α

on a small neighborhood around ϕα(p); this map is smooth since the above charts areall smoothly compatible.

Theorem 11. [Exercise 2.7(2)] Let M and N be smooth manifolds with or withoutboundary, and let F : M → N be a map.

(1) If every point p ∈ M has a neighborhood U such that the restriction F |U issmooth, then F is smooth.

(2) Conversely, if F is smooth, then its restriction to every open subset is smooth.

4

Proof. (1) is obvious. For (2), let E be an open subset of M . Let p ∈ E and choosesmooth charts (U,ϕ) containing p and (V, ψ) containing F (p) such that F (U) ⊆ V andψ F ϕ−1 is smooth. By restricting ϕ to U ∩E, we find that F |E is also smooth.

Theorem 12. [Exercise 2.9] Suppose F : M → N is a smooth map between smoothmanifolds with or without boundary. Then the coordinate representation of F withrespect to every pair of smooth charts for M and N is smooth.

Proof. Almost identical to Theorem 9.

Theorem 13. [Exercise 2.11] Let M , N and P be smooth manifolds with or withoutboundary.

(1) Every constant map c : M → N is smooth.(2) The identity map of M is smooth.(3) If U ⊆ M is an open submanifold with or without boundary, then the inclusion

map U →M is smooth.

Proof. If M 6= ∅ then let y be the constant value that c assumes and let (V, ψ) be asmooth chart containing y. If x ∈M and (U,ϕ) is any smooth chart containing x thenψ c ϕ−1 is constant and therefore smooth. This proves (1). Part (2) follows from thefact that any two coordinate maps for M are smoothly compatible. For (3), if p ∈ Uand (E,ϕ) is any smooth chart in U containing p then ϕ ϕ−1 = Idϕ(E) is smooth.

Theorem 14. [Exercise 2.16]

(1) Every composition of diffeomorphisms is a diffeomorphism.(2) Every finite product of diffeomorphisms between smooth manifolds is a diffeo-

morphism.(3) Every diffeomorphism is a homeomorphism and an open map.(4) The restriction of a diffeomorphism to an open submanifold with or without

boundary is a diffeomorphism onto its image.(5) “Diffeomorphic” is an equivalence relation on the class of all smooth manifolds

with or without boundary.

Proof. If f : M → N and g : N → P are diffeomorphisms then f−1 g−1 is a smoothinverse to g f , which proves (1). If fi : Mi → Ni are diffeomorphisms for i = 1, . . . , nthen f−1

1 ×· · ·×f−1n is a smooth inverse to f1×· · ·×fn, which proves (2). Part (3) follows

from the fact that smooth maps are continuous, and part (4) follows immediately from(3) and Theorem 11. Part (5) follows from (1).

Theorem 15. [Exercise 2.19] Suppose M and N are smooth manifolds with boundaryand F : M → N is a diffeomorphism. Then F (∂M) = ∂N , and F restricts to adiffeomorphism from IntM to IntN .

5

Proof. It suffices to show that F (∂M) ⊆ ∂N , for then F (IntM) = IntN by Theorem1.46 and F |IntM is a diffeomorphism by Theorem 14. Let x ∈ ∂M and choose smoothcharts (U,ϕ) containing x and (V, ψ) containing F (x) such that ϕ(x) ∈ ∂Hn, F (U) = Vand ψ F ϕ−1 is a diffeomorphism. Then ϕF−1 is a smooth coordinate map sendingF (x) to ϕ(x) ∈ ∂Hn, so F (x) ∈ ∂N by definition.

Example 16. [Exercise 2.27] Give a counterexample to show that the conclusion ofthe extension lemma can be false if A is not closed.

Take f(x) = 1/x defined on R \ 0.

Example 17. [Problem 2-1] Define f : R→ R by

f(x) =

1, x ≥ 0,

0, x < 0.

Show that for every x ∈ R, there are smooth coordinate charts (U,ϕ) containing x and(V, ψ) containing f(x) such that ψ f ϕ−1 is smooth as a map from ϕ (U ∩ f−1(V ))to ψ(V ), but f is not smooth as defined in this chapter.

f is clearly smooth on R\0. If x = 0 then ϕ = Id(−1,1) is a coordinate map containingx and ψ = Id(1/2,3/2) is a coordinate map containing f(x) = 1 such that ψ f ϕ−1 issmooth on ϕ ((−1, 1) ∩ f−1 ((1/2, 3/2))) = ϕ ([0, 1)) = [0, 1). However, f is clearly notsmooth since it is not continuous at 0.

Theorem 18. Let M1, . . . ,Mk−1 be smooth manifolds without boundary and let Mk bea smooth manifold with boundary. Then each projection map πi : M1 × · · · ×Mk →Mi

is smooth.

Proof. Let ni be the dimension of Mi. Let x = (x1, . . . , xk) ∈M1 × · · · ×Mk. For eachi = 1, . . . , k, choose a smooth chart (Ui, ϕi) containing xi. Write U = U1×· · ·×Uk andϕ = ϕ1 × · · · × ϕk. Then (U,ϕ) is a smooth chart containing x and ϕi πi ϕ−1 = piis smooth, where pi is a projection from Rn1 × · · · ×Hnk to Rni (or Hni if i = k). Thisshows that πi is smooth.

Theorem 19. [Problem 2-2] Suppose M1, . . . ,Mk and N are smooth manifolds withor without boundary, such that at most one of M1, . . . ,Mk has nonempty boundary.For each i, let πi : M1 × · · · ×Mk → Mi denote the projection onto the Mi factor. Amap F : N → M1 × · · · ×Mk is smooth if and only if each of the component mapsFi = πi F : N →Mi is smooth.

Proof. For simplicity we assume that Mk is a smooth manifold with boundary. FromProposition 2.10(d) it is clear that every Fi is smooth if F is smooth. Suppose thateach Fi is smooth and let x ∈ N . For each i, choose smooth charts (Ui, ϕi) containing xand (Vi, ψi) containing Fi(x) such that ψi Fi ϕ−1

i = ψi πi F ϕ−1i . Replace U1 with

6

U = U1 ∩ · · · ∩Uk and replace ϕ1 with ϕ1|U ; by Theorem 12, each map ψi πi F ϕ−11

is smooth. Write V = V1 × · · · × Vk and ψ = ψ1 × · · · × ψk so that (V, ψ) is a smoothchart containing F (x). Since the ith component of ψ F ϕ−1

1 is just ψi πi F ϕ−11 ,

the map ψ F ϕ−11 is smooth. This shows that F is smooth.

Theorem 20. [Problem 2-4] The inclusion map f : Bn → Rn is smooth when Bn isregarded as a smooth manifold with boundary.

Proof. It is clear that f is smooth on IntBn = Bn. For points on ∂Bn, the map providedby Problem 3-4 in [[1]] is a suitable coordinate map.

Theorem 21. [Problem 2-7] Let M be a nonempty smooth n-manifold with or withoutboundary, and suppose n ≥ 1. Then the vector space C∞(M) is infinite-dimensional.

Proof. We first show that any collection fα from C∞(M) with nonempty disjointsupports is linearly independent. Suppose there exist r1, . . . , rk ∈ R such that

r1fα1 + · · ·+ rkfαk = 0

where α1, . . . , αk are distinct. For each i, choose some xi ∈ supp fαi such that fαi(xi) 6=0. Then

rifαi(xi) = (r1fα1 + · · ·+ rkfαk)(xi) = 0,

so ri = 0 since fαi(xi) 6= 0.

Choose some coordinate cube U ⊆M and a homeomorphism ϕ : U → (0, 1)n. For eachi = 1, 2, . . . , let Bi = (0, 1)n−1 × (1/(i + 1), 1/i) and let Ai be any nonempty closedsubset of Bi. By Proposition 2.25, there is a smooth bump function fi for ϕ−1(Ai)supported in ϕ−1(Bi). Furthermore, since the sets ϕ−1(Bi) are disjoint, the functionsfi are linearly independent. This shows that C∞(M) is infinite-dimensional.

Theorem 22. [Problem 2-8] Define F : Rn → RPn by F (x1, . . . , xn) = [x1, . . . , xn, 1].Then F is a diffeomorphism onto a dense open subset of RPn. A similar statementholds for G : Cn → CPn defined by G(z1, . . . , zn) = [z1, . . . , zn, 1].

Proof. The inverse to F is given by

F−1[y1, . . . , yn, yn+1] = (y1/yn+1, . . . , yn/yn+1);

it is easy to check that the two maps are smooth since F−1 itself is a coordinate mapfor RPn. The image of F is dense in RPn since any point [x1, . . . , xn, 0] not in the imageof F can be approximated by some sequence [x1, . . . , xn, ak] where ak → 0.

Theorem 23. [Problem 2-9] Given a polynomial p in one variable with complex coef-ficients, not identically zero, there is a unique smooth map p : CP1 → CP1 that makesthe following diagram commute, where CP1 is 1-dimensional complex projective spaceand G : C→ CP1 is the map of Theorem 22:

7

C CP1

C CP1

G

p p

G

Proof. We can assume that p is not constant, for otherwise the result is clear. For anyz = [z1, z2] ∈ CP1, define

p(z) =

[p(z1/z2), 1], z2 6= 0,

[1, 0], z2 = 0.

The map p clearly satisfies the given diagram and is smooth on the image of G, so itremains to show that p is smooth in a neighborhood of [1, 0]. Define a diffeomorphismH : C→ CP1 by

H(z) = [1, z];

by definition, (H(C), H−1) is a smooth chart for CP1. We can compute

(H−1 p H)(z) = (H−1 p)[1, z]

=

H−1[p(1/z), 1], z 6= 0,

H−1[1, 0], z = 0,

=

1/p(1/z), z 6= 0,

0, z = 0,

and it is easy to check that this map is smooth in a neighborhood of 0 wheneverp is non-constant (since 1/p(1/z) is a rational function in z). This shows that p issmooth.

Theorem 24. [Problem 2-10] For any topological space M , let C(M) denote the algebraof continuous functions f : M → R. Given a continuous map F : M → N , defineF ∗ : C(N)→ C(M) by F ∗(f) = f F .

(1) F ∗ is a linear map.(2) Suppose M and N are smooth manifolds. Then F : M → N is smooth if and

only if F ∗(C∞(N)) ⊆ C∞(M).(3) Suppose F : M → N is a homeomorphism between smooth manifolds. Then it

is a diffeomorphism if and only if F ∗ restricts to an isomorphism from C∞(N)to C∞(M).

8

Proof. (1) is obvious. For (2), if F is smooth and f ∈ C∞(N) then F ∗(f) = f F ∈C∞(M) by Proposition 2.10(d). Conversely, suppose that F ∗(C∞(N)) ⊆ C∞(M). Letx ∈ M and let n be the dimension of N . Choose smooth charts (U,ϕ) containingx and (V, ψ) containing F (x) such that F (U) ⊆ V . For each i = 1, . . . , n the mapπi ψ F = F ∗(πi ψ) is smooth, where πi : Rn → R is the projection onto the ithcoordinate. By 9, πi ψ F ϕ−1 is smooth for i = 1, . . . , n, so ψ F ϕ−1 must besmooth. This shows that F is smooth. Part (3) follows directly from part (2): if Fis a diffeomorphism then (F−1)∗ : C∞(M) → C∞(N) is clearly an inverse to F ∗, andconversely if F ∗|C∞(N) is an isomorphism then F and F−1 are both smooth.

Theorem 25. [Problem 2-11] Suppose V is a real vector space of dimension n ≥ 1.Define the projectivization of V , denoted by P(V ), to be the set of 1-dimensionallinear subspaces of V , with the quotient topology induced by the map π : V \0 → P(V )that sends x to its span. Then P(V ) is a topological (n− 1)-manifold, and has a uniquesmooth structure with the property that for each basis (E1, . . . , En) for V , the mapE : RPn−1 → P(V ) defined by E[v1, . . . , vn] = [viEi] is a diffeomorphism.

Proof. Obvious.

Theorem 26. [Problem 2-14] Suppose A and B are disjoint closed subsets of a smoothmanifold M . There exists f ∈ C∞(M) such that 0 ≤ f(x) ≤ 1 for all x ∈ M ,f−1(0) = A, and f−1(1) = B.

Proof. By Theorem 2.29, there are functions fA, fB : M → R such that f−1A (0) = A

and f−1B (0) = B; the function defined by

f(x) =fA(x)

fA(x) + fB(x)

satisfies the required properties.

Chapter 3. Tangent Vectors

Theorem 27. [Exercise 3.7] Let M , N and P be smooth manifolds with or withoutboundary, let F : M → N and G : N → P be smooth maps, and let p ∈M .

(1) dFp : TpM → TF (p)N is linear.(2) d(G F )p = dGF (p) dFp : TpM → TGF (p)P .(3) d(IdM)p = IdTpM : TpM → TpM .(4) If F is a diffeomorphism, then dFp : TpM → TF (p)N is an isomorphism, and

(dFp)−1 = d(F−1)F (p).

9

Proof. If v, w ∈ TpM and c ∈ R then

dFp(v + w)(f) = (v + w)(f F ) = v(f F ) + w(f F ) = dFp(v)(f) + dFp(w)(f)

and

dFp(cv)(f) = (cv)(f F ) = cdFp(v)(f)

for all f ∈ C∞(N), which proves (1). To prove (2), we have

d(G F )p(v)(f) = v(f G F ) = dFp(v)(f G) = dGF (p)(dFp(v))(f)

for all f ∈ C∞(N) and v ∈ TpM , so d(G F )p = dGF (p) dFp. For (3),

d(IdM)p(v)(f) = v(f IdM) = v(f),

so d(IdM)p(v) = v for all v ∈ TpM and therefore d(IdM)p = IdTpM . Part (4) followsimmediately, since

d(F−1)F (p) dFp = d(F−1 F )p = d(IdM)p = IdTpM

and similarly dFp d(F−1)F (p) = IdTF (p)N .

Example 28. [Exercise 3.17] Let (x, y) denote the standard coordinates on R2. Verifythat (x, y) are global smooth coordinates on R2, where

x = x, y = y + x3.

Let p be the point (1, 0) ∈ R2 (in standard coordinates), and show that

∂

∂x

∣∣∣∣p

6= ∂

∂x

∣∣∣∣p

,

even though the coordinate functions x and x are identically equal.

Since an inverse to (x, y) 7→ (x, y+ x3) is given by (x, y) 7→ (x, y− x3), the coordinatesare smooth and global. Now

∂

∂x

∣∣∣∣p

=∂

∂x

∣∣∣∣p

+ 3∂

∂y

∣∣∣∣p

,

∂

∂y

∣∣∣∣p

=∂

∂y

∣∣∣∣p

,

so ∂/∂x|p 6= ∂/∂x|p.

Theorem 29. Let X be a connected topological space and let ∼ be an equivalencerelation on X. If every x ∈ X has a neighborhood U such that p ∼ q for every p, q ∈ U ,then p ∼ q for every p, q ∈ X.

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Proof. Let p ∈ X and let S = q ∈ X : p ∼ q. If q ∈ S then there is a neighborhoodU of q such that q1 ∼ q2 for every q1, q2 ∈ U . In particular, for every r ∈ U we havep ∼ q and q ∼ r which implies that p ∼ r, and U ⊆ S. This shows that S is open. Ifq ∈ X \S then there is a neighborhood U of q such that q1 ∼ q2 for every q1, q2 ∈ U . Ifp ∼ r for some r ∈ U then p ∼ q since q ∼ r, which contradicts the fact that q ∈ X \S.Therefore U ⊆ X \ S, which shows that S is closed. Since X is connected, S = X.

Corollary 30. Let X be a connected topological space and let f : X → Y be a con-tinuous map. If every x ∈ X has a neighborhood U on which f is constant, then f isconstant on X.

Theorem 31. [Problem 3-1] Suppose M and N are smooth manifolds with or withoutboundary, and F : M → N is a smooth map. Then dFp : TpM → TF (p)N is the zeromap for each p ∈M if and only if F is constant on each component of M .

Proof. Suppose F is constant on each component of M and let p ∈M . For all v ∈ TpMwe have dFp(v)(f) = v(f F ) = 0 since f F is constant on a neighborhood of p (seeProposition 3.8), so dFp = 0. Conversely, suppose that dF = 0, let p ∈ M , and choosesmooth charts (U,ϕ) containing p and (V, ψ) containing F (p) such that ϕ(U) is an open

ball and F (U) ⊆ V . Let F = ψ F ϕ−1 be the coordinate representation of F . From

equation (3.10), the derivative of F is zero since dFp = 0 and therefore F is constant

on ϕ(U). This implies that F = ψ−1 F ϕ is constant on U . By Corollary 30, F isconstant on each component of M .

Theorem 32. [Problem 3-2] Let M1, . . . ,Mk be smooth manifolds, and for each j,let πj : M1 × · · · × Mk → Mj be the projection onto the Mj factor. For any pointp = (p1, . . . , pk) ∈M1 × · · · ×Mk, the map

α : Tp(M1 × · · · ×Mk)→ Tp1M1 ⊕ · · · ⊕ TpkMk

defined by

α(v) = (d(π1)p(v), . . . , d(πk)p(v))

is an isomorphism. The same is true if one of the spaces Mi is a smooth manifold withboundary.

Proof. It is clear that α is a linear map. For each j = 1, . . . , k, define ej : Mj →M1 × · · · ×Mk by ej(x) = (p1, . . . , x, . . . , pk) where x is in the jth position. Each ejinduces a linear map d(ej)pj : TpjMj → Tp(M1 × · · · ×Mk), so we can define a linearmap

β : Tp1M1 ⊕ · · · ⊕ TpkMk → Tp(M1 × · · · ×Mk)

(v1, . . . , vk) 7→ d(e1)p1(v1) + · · ·+ d(ek)pk(vk).

11

The jth component of (α β)(v1, . . . , vk) is then given by

d(πj)p

(k∑i=1

d(ei)pi(vi)

)=

k∑i=1

d(πj ei)pi(vi)

= d(πj ej)pj(vj)= vj

since πj ei is constant when i 6= j, and πj ej = IdMj. This shows that α is surjective.

But

dim(Tp(M1 × · · · ×Mk)) = dim(Tp1M1 ⊕ · · · ⊕ TpkMk),

so α is an isomorphism.

Theorem 33. If (Uj, ϕj) are smooth charts containing pj as defined in Theorem 32,then ϕ = ϕ1×· · ·×ϕk is a smooth coordinate map containing p. Let nj be the dimensionof Mj, and write ϕj = (x1

j , . . . , xnjj ) for each j so that

ϕ = ϕ1 × · · · × ϕk = (x11, . . . , x

n11 , . . . , x

1k, . . . , x

nkk )

where xij = xij πj.

(1) The collection ∂

∂xij

∣∣∣∣p

: 1 ≤ j ≤ k, 1 ≤ i ≤ nj

forms a basis for Tp(M1 × · · · ×Mk).

(2) Let v ∈ Tp(M1 × · · · × Mk) and write v =∑

i,j vji ∂/∂xij

∣∣p. Then the jth

component of α(v) is∑

i vji ∂/∂xij

∣∣pj

(the variable j is fixed).

Proof. Since Bj =∂/∂x1

j

∣∣pj, . . . , ∂/∂x

njj

∣∣pj

is a basis for TpjMj, the set α−1(B1 ∪

· · · ∪ Bk) = β(B1 ∪ · · · ∪ Bk) is a basis for Tp(M1 × · · · ×Mk). But

β

(∂

∂xij

∣∣∣∣pj

)(f) = d(ej)pj

(∂

∂xij

∣∣∣∣pj

)(f)

= d(ej ϕ−1j )ϕj(pj)

(∂

∂xij

∣∣∣∣ϕj(pj)

)(f)

=∂(f ej ϕ−1

j )

∂xij(ϕj(pj))

=∂(f ϕ−1)

∂xij(ϕ(p))

12

= d(ϕ−1)ϕ(p)

(∂

∂xij

∣∣∣∣ϕ(p)

)(f)

=∂

∂xij

∣∣∣∣p

(f),

which proves (1). For (2), we compute the jth component of α(v) as

d(πj)p

(∑r,s

vsr∂

∂xrs

∣∣∣∣p

)(f) =

∑r,s

vsrd(πj)p

(∂

∂xrs

∣∣∣∣p

)(f)

=∑r

vjr∂(f πj ϕ−1)

∂xrj(ϕ(p))

=∑r

vjr∂(f ϕ−1

j )

∂xrj(ϕj(pj))

=∑i

vji

(∂

∂xij

∣∣∣∣pj

)(f).

Theorem 34. [Problem 3-3] If M and N are smooth manifolds, then T (M × N) isdiffeomorphic to TM × TN .

Proof. Let m and n denote the dimensions of M and N respectively. For each (p, q) ∈M × N , there is an isomorphism α(p,q) : T(p,q)(M × N) → TpM ⊕ TqN as in Theorem32. Define a bijection F : T (M × N) → TM × TN by sending each tangent vectorv ∈ T(p,q)(M × N) to α(p,q)(v). For any (p, q) ∈ M × N , choose smooth charts (U,ϕ)containing p and (V, ψ) containing q. Let π : T (M × N) → M × N be the naturalprojection map. Then ρ = ϕ× ψ is a smooth coordinate map for M ×N , ρ : π−1(U ×V )→ R2(m+n) is a smooth coordinate map for T (M×N), and ϕ×ψ : π−1(U)×π−1(V )→R2(m+n) is a smooth coordinate map for TM × TN . By Theorem 33 the coordinaterepresentation of F is

F (x1, . . . , xm, y1, . . . , yn, v1, . . . , vm+n)

= (x1, . . . , xm, v1, . . . , vm, y1, . . . , yn, vm+1, . . . , vm+n)

which is clearly smooth. Similarly, F−1 is also smooth.

Theorem 35. [Problem 3-4] TS1 is diffeomorphic to S1 × R.

Proof. We begin by showing that for each z ∈ S1 we can choose an associated tangentvector vz ∈ TzS1. Let θ : U → R be any angle function defined on a neighborhood of

13

z, and set vz = d/dx|z = (dθ)−1z d/dx|θ(z). If θ′ : U ′ → R is any other angle function

defined around z then θ − θ′ is constant on U ∩ U ′ and (θ′ θ−1)(t) = t + c for somec ∈ R. Therefore (dθ′)−1

z d/dx|θ′(z) = (dθ′)−1z d/dx′|θ′(z), and vz is well-defined.

Define a map F : S1 × R → TS1 by sending (z, r) to rvz. If θ : U → R is an anglefunction defined around z then the coordinate representation of F is F (z, r) = (θ(z), r),so F is smooth. An inverse to F is defined by sending v ∈ TzS1 to (z, r), where v = rvz(since TS1 is one-dimensional). The coordinate representation of F−1 is F−1(x, r) =(θ−1(x), r), so F−1 is also smooth.

Theorem 36. [Problem 3-5] Let S1 ⊆ R2 be the unit circle, and let K ⊆ R2 be theboundary of the square of side 2 centered at the origin: K = (x, y) : max(|x| , |y|) = 1.There is a homeomorphism F : R2 → R2 such that F (S1) = K, but there is no diffeo-morphism with the same property.

Proof. The map defined by

F (x, y) =

√x2 + y2

max(|x| , |y|)(x, y)

is a homeomorphism such that F (S1) = K. Suppose there is a diffeomorphism F : R2 →R2 such that F (S1) = K. Let γ : (−1, 1)→ S1 be given by s 7→ exp(2πi(s+ c)), wherec is chosen so that (F γ)(0) = (1, 1). We can assume without loss of generality that(F γ)((−ε, 0)) lies on the right edge of K and that (F γ)((0, ε)) lies on the top edge ofK, for small ε. Then there are smooth functions γ1, γ2 such that (F γ)(t) = (1, γ1(t))for t ∈ (−ε, 0) and (F γ)(t) = (γ2(t), 1) for t ∈ (0, ε). Since (F γ)′(t) = (0, γ′1(t))and (F γ)′(t) = (γ′2(t), 0), taking t→ 0 shows that (F γ)′(0) = 0. But dF (γ′(0)) 6= 0since γ′(0) 6= 0 and dF is an isomorphism, so this is a contradiction.

Theorem 37. [Problem 3-7] Let M be a smooth manifold with or without boundaryand p be a point of M . Let C∞p (M) denote the algebra of germs of smooth real-valuedfunctions at p, and let DpM denote the vector space of derivations of C∞p (M). Definea map φ : DpM → TpM by (φv)f = v([f ]p). Then φ is an isomorphism.

Proof. It is clear that φ is a homomorphism. If φv = 0 then v([f ]p) = 0 for allf ∈ C∞(M), so v = 0 and therefore φ is injective. If v ∈ TpM then we can definev′ ∈ DpM by setting v′([f ]p) = v(f). This is well-defined due to Proposition 3.8. Thenφv′ = v, so φ is surjective.

Theorem 38. [Problem 3-8] Let M be a smooth manifold with or without boundaryand p ∈ M . Let VpM denote the set of equivalence classes of smooth curves startingat p under the relation γ1 ∼ γ2 if (f γ1)′(0) = (f γ2)′(0) for every smooth real-valued function f defined in a neighborhood of p. The map ψ : VpM → TpM defined byψ[γ] = γ′(0) is well-defined and bijective.

14

Proof. If γ1, γ2 ∈ [γ] then γ′1(0)(f) = (f γ1)′(0) = (f γ2)′(0) = γ′2(0)(f), whichshows that ψ is well-defined. The same argument shows that ψ is injective. Finally,Proposition 3.23 shows that ψ is surjective.

Chapter 4. Submersions, Immersions, and Embeddings

Theorem 39. [Exercise 4.4]

(1) The composition of two smooth submersions is a smooth submersion.(2) The composition of two smooth immersions is a smooth immersion.(3) The composition of two maps of constant rank need not have constant rank.

Proof. Let F : M → N and G : N → P be smooth maps; parts (1) and (2) follow fromthe fact that d(G F )x = dGF (x) dFx. For (3), define F : R→ R2 and G : R2 → R by

F (x) = (1, x), G(x, y) = x+ y2

so that

DF (x) =

[01

], DG(x, y) =

[1 2y

], D(G F )(x) = 2x.

Theorem 40. [Exercise 4.7]

(1) Every composition of local diffeomorphisms is a local diffeomorphism.(2) Every finite product of local diffeomorphisms between smooth manifolds is a local

diffeomorphism.(3) Every local diffeomorphism is a local homeomorphism and an open map.(4) The restriction of a local diffeomorphism to an open submanifold with or without

boundary is a local diffeomorphism.(5) Every diffeomorphism is a local diffeomorphism.(6) Every bijective local diffeomorphism is a diffeomorphism.(7) A map between smooth manifolds with or without boundary is a local diffeomor-

phism if and only if in a neighborhood of each point of its domain, it has acoordinate representation that is a local diffeomorphism.

Proof. Let F : M → N and G : N → P be local diffeomorphisms. If x ∈M then thereare neighborhoods U of x and V of F (x) such that F |U : U → V is a diffeomorphism,and there are neighborhoods V ′ of F (x) and W of (G F )(x) such that G|V ′ : V ′ →W is a diffeomorphism. Then (G F )|F−1(V ∩V ′) is a diffeomorphism onto its image,which proves (1). Parts (2)-(5) are obvious. Suppose F : M → N is a bijective localdiffeomorphism. If y ∈ N then there are neighborhoods U of F−1(y) and V of y suchthat F |U : U → V is a diffeomorphism, so (F−1)|V is smooth. Theorem 11 then showsthat F−1 is smooth, which proves (6). Part (7) is obvious.

15

Theorem 41. [Exercise 4.10] Suppose M , N , P are smooth manifolds with or withoutboundary, and F : M → N is a local diffeomorphism.

(1) If G : P →M is continuous, then G is smooth if and only if F G is smooth.(2) If in addition F is surjective and G : N → P is any map, then G is smooth if

and only if G F is smooth.

Proof. For (1), if F G is smooth and x ∈ P then there are neighborhoods U of G(x)and V of (F G)(x) such that F |U : U → V is a diffeomorphism, so G|G−1(U) =(F |U)−1 (F G)|G−1(U) is smooth. By Theorem 11, G is smooth. For (2), if G Fis smooth and x ∈ N then there is a point p ∈ M such that x = F (p), and there areneighborhoods U of p and V of x such that F |U : U → V is a diffeomorphism. ThenG|V = G F (F |U)−1 is smooth, so G is smooth by Theorem 11.

Example 42. [Exercise 4.24] Give an example of a smooth embedding that is neitheran open nor a closed map.

Let X = [0, 1) and Y = [−1, 1], and let f : X → Y be the identity map on X. Thenf is a smooth embedding, X is both open and closed, but f(X) is neither open norclosed.

Example 43. [Exercise 4.27] Give an example of a smooth map that is a topologicalsubmersion but not a smooth submersion.

Consider f(x) = x3 at x = 0.

Theorem 44. [Exercise 4.32] Suppose that M , N1, and N2 are smooth manifolds, andπ1 : M → N1 and π2 : M → N2 are surjective submersions that are constant oneach other’s fibers. Then there exists a unique diffeomorphism F : N1 → N2 such thatF π1 = π2.

Proof. Theorem 4.30 shows that there are unique smooth maps π1 : N2 → N1 andπ2 : N1 → N2 such that π1 = π1 π2 and π2 = π2 π1. Then π2 π1 π2 = π2 π1 = π2,so π2 π1 = IdN2 by the uniqueness part of Theorem 4.30. Similarly, π1 π2 = IdN1 .

Theorem 45. [Exercise 4.33]

(1) Every smooth covering map is a local diffeomorphism, a smooth submersion, anopen map, and a quotient map.

(2) An injective smooth covering map is a diffeomorphism.(3) A topological covering map is a smooth covering map if and only if it is a local

diffeomorphism.

Proof. Let π : E →M be a smooth covering map. Let e ∈ E and x = π(e), let U be an

evenly covered neighborhood of x, and let U be the component of π−1(U) containing

16

e. By definition π|U : U → U is a diffeomorphism, so π is a local diffeomorphism andthe other properties in (1) follow. If π is injective then π is a bijection. Every bijectivelocal diffeomorphism is a diffeomorphism, proving (2). Part (3) is obvious.

Theorem 46. [Exercise 4.37] Suppose π : E → M is a smooth covering map. Everylocal section of π is smooth.

Proof. Let τ : V → E be a local section of π where V is open in M . Let x ∈ V and letU be an evenly covered neighborhood of x contained in V . By Proposition 4.36, thereis a unique smooth local section σ : U → E such that σ(x) = τ(x). Therefore τ |U = σ,and τ is smooth by Theorem 11.

Theorem 47. [Exercise 4.38] Suppose E1, . . . , Ek and M1, . . . ,Mk are smooth man-ifolds (without boundary), and πi : Ei → Mi is a smooth covering map for eachi = 1, . . . , k. Then π1 × · · · × πk : E1 × · · · × Ek → M1 × · · · × Mk is a smoothcovering map.

Proof. This is true for topological covering maps. But any finite product of local dif-feomorphisms is a local diffeomorphism, so the result follows from Theorem 45.

Theorem 48. Suppose π : E →M is a smooth covering map. If U ⊆M is evenly cov-ered in the topological sense, then π maps each component of π−1(U) diffeomorphicallyonto U .

Proof. Let U be a component of π−1(U) and suppose π|U : U → U is a homeomorphism(or merely a bijection). Since π|U is a local diffeomorphism and it is bijective, it is adiffeomorphism.

Theorem 49. [Exercise 4.42] Suppose M is a connected smooth n-manifold with bound-ary, and π : E →M is a topological covering map. Then E is a topological n-manifoldwith boundary such that ∂E = π−1(∂M), and it has a unique smooth structure suchthat π is a smooth covering map.

Proof. Identical to Proposition 4.40.

Theorem 50. [Exercise 4.44] If M is a connected smooth manifold, there exists a

simply connected smooth manifold M , called the universal covering manifold of M , and

a smooth covering map π : M → M . The universal covering manifold is unique in the

following sense: if M ′ is any other simply connected smooth manifold that admits a

smooth covering map π′ : M ′ → M , then there exists a diffeomorphism Φ : M → M ′

such that π′ Φ = π.

17

Proof. Every manifold has a universal covering space, so let π : M →M be a universal

covering. Proposition 4.40 shows that M is a smooth manifold with a unique smooth

structure such that π is a smooth covering map. If π′ : M ′ → M is another universal

smooth covering, then there is a homeomorphism Φ : M → M ′ such that π′ Φ =

π. Let e ∈ M , let x = π(e) and choose some e′ ∈ π−1(x). Let U be an evenly

covered neighborhood of x with respect to both π and π′, let U be the component of

π−1(U) containing e, and let U ′ be the component of (π′)−1(U) containing e′. ThenΦ|U = (π′|U ′)−1 π|U , so Φ|U is a diffeomorphism. This shows that Φ is a bijective localdiffeomorphism and therefore Φ is a diffeomorphism.

Example 51. [Problem 4-1] Use the inclusion map ι : Hn → Rn to show that Theorem4.5 does not extend to the case in which M is a manifold with boundary.

Lemma 3.11 shows that dι0 is invertible. If Theorem 4.5 holds, then there are neigh-borhoods U ⊆ Hn of 0 and a neighborhood V ⊆ Rn of 0 such that ι|U → V is adiffeomorphism. Since ι is a bijection, we must have U = V . This is impossible,because U cannot be open in Rn.

Theorem 52. [Problem 4-2] Suppose M is a smooth manifold (without boundary), Nis a smooth manifold with boundary, and F : M → N is smooth. If p ∈ M is a pointsuch that dFp is nonsingular, then F (p) ∈ IntN .

Proof. Let n be the dimension of M , which is the same as the dimension of N . SupposeF (p) ∈ ∂N . Choose smooth charts (U,ϕ) centered at p and (V, ψ) centered at F (p)

such that ψ(V ) ⊆ Hn, F (U) ⊆ V and ψ(F (p)) = 0. Let F = ψ F ϕ−1. Then dF0

is invertible, so the inverse function theorem shows that there are neighborhoods U0 of

0 and V0 of 0 open in Rn such that F |U0: U0 → V0 is a diffeomorphism. But V0 is a

neighborhood of 0 open in Rn and contained in Hn, which is impossible.

Example 53. [Problem 4-4] Let γ : R→ T2 be the curve of Example 4.20. The imageset γ(R) is dense in T2.

Let (z1, z2) ∈ T2 and choose s1, s2 ∈ R so that e2πis1 = z1 and e2πis2 = z2. Thenγ(s1 + n) = (e2πis1 , e2πiαs1e2πiαn), so it remains to show that

S =e2πiαn : n ∈ Z

is dense in S1 whenever α is irrational. Let ε > 0. By Lemma 4.21, there exist integersn,m such that |nα−m| < ε. Let θ = nα−m so that

e2πiθ = e2πiαn ∈ Sand θ 6= 0 since α is irrational. Any integer power of e2πiθ is also a member of S, andsince ε was arbitrary, we can approximate any point on S1 by taking powers of e2πiθ.

18

Example 54. We identify S3 with the set

(z, w) ∈ C2 : |z|2 + |w|2 = 1

. Let p : C→S2 \ N be the inverse of the stereographic projection given by

p(x+ yi) =1

x2 + y2 + 1(2x, 2y, x2 + y2 − 1),

or

p(z) =1

|z|2 + 1(z + z,−i(z − z), |z|2 − 1).

Define q : S3 \ S1 × 0 → C by (z, w) = z/w. Then we have a smooth mapp q : S3 \ S1 × 0 → S2 \ N given by

(p q)(z, w) =1

|z/w|2 + 1

(z

w+z

w,−i

( zw− z

w

),∣∣∣ zw

∣∣∣2 − 1

)= (zw + zw,−i(zw − zw), zz − ww).

It is easy to see that we can extend p q so that it maps S3 into S2 by using the formulaabove.

Theorem 55. [Problem 4-5] Let CPn be n-dimensional complex projective space.

(1) The quotient map π : Cn+1 \ 0 → CPn is a surjective smooth submersion.(2) CP1 is diffeomorphic to S2.

Proof. Let (z1, . . . , zn+1) ∈ Cn+1\0 and assume without loss of generality that zn+1 6=0. We have a coordinate map ϕ given by

[w1, . . . , wn, wn+1] 7→(

w1

wn+1

, . . . ,wnwn+1

).

Then

(ϕ π)′(w1, . . . , wn+1) =

w−1n+1 0 · · · 0 −w1w

−2n+1

0 w−1n+1 · · · 0 −w2w

−2n+1

......

. . ....

...0 0 · · · w−1

n+1 −wnw−2n+1

,which clearly has full rank when wk = zk for k = 1, . . . , n+ 1 (since the first n columnsform an invertible matrix). This proves (1). The map π : C2\0 → CP1 is a surjectivesmooth submersion. Define a map ψ : C2 \ 0 → S2 by

ψ(z1, z2) = (p q)(

(z1, z2)

|(z1, z2)|

),

where p q is the map in 54. By Theorem 4.30, ψ descends to a unique smooth map

ψ : CP1 → S2 satisfying ψ π = ψ. This map is a diffeomorphism.

Theorem 56. [Problem 4-6] Let M be a nonempty smooth compact manifold. Thereis no smooth submersion F : M → Rk for any k > 0.

19

Proof. By Proposition 4.28, F (M) is open in Rk. Also, F (M) is compact because M iscompact. Therefore F (M) is open, closed and bounded. But Rk is connected and Mis nonempty, so F (M) = Rk. This contradicts the boundedness of F (M).

Theorem 57. [Problem 4-7] Suppose M and N are smooth manifolds, and π : M → Nis a surjective smooth submersion. There is no other smooth manifold structure on

N that satisfies the conclusion of Theorem 4.29; in other words, assuming that Nrepresents the same set as N with a possibly different topology and smooth structure,

and that for every smooth manifold P with or without boundary, a map F : N → P is

smooth if and only if F π is smooth, then IdN is a diffeomorphism between N and N .

Proof. Write ιN for the identity map from N to N , and ιN for the identity map from

N to N . Then ιN is smooth if and only if ιN π = π is smooth and ιN is smooth if andonly if ιN π = π is smooth. Both of these conditions hold, and since ιN and ιN areinverses, IdN = ιN = ιN is a diffeomorphism.

Theorem 58. [Problem 4-8] Let π : R2 → R be defined by π(x, y) = xy. Then π issurjective and smooth, and for each smooth manifold P , a map F : R → P is smoothif and only if F π is smooth; but π is not a smooth submersion (cf. Theorem 4.29).

Proof. It is clear that F π is smooth if F is smooth. If F π is smooth then themap x 7→ (F π)(x, 1) is smooth and identical to F , so F is smooth. However, we cancompute

Dπ(x, y) =[y x

],

which does not have full rank when (x, y) = (0, 0).

Theorem 59. [Problem 4-9] Let M be a connected smooth manifold and let π : E →Mbe a topological covering map. There is only one smooth structure on E such that π isa smooth covering map (cf. Proposition 4.40).

Proof. LetM be the smooth structure constructed in Proposition 4.40. Suppose thereis another smooth structureM′ for which π is a smooth covering map. Let p ∈ E and let(U, ϕ) be a smooth chart inM′ containing p. Let V be an evenly covered neighborhood

of ϕ(p) and let V be the component of π−1(V ) containing p so that π|V : V → V is

a diffeomorphism (with respect to M′). By shrinking V and V if necessary, we mayassume that there is a smooth chart (V, ψ) containing ϕ(p). Then ψ π|V is a smoothcoordinate map in M, and

ϕ (ψ π|V )−1 = ϕ (π|V )−1 ψ−1

is smooth. This shows that ϕ and ψ π|V are smoothly compatible, and since p wasarbitrary, M =M′.

20

Example 60. [Problem 4-10] Show that the map q : Sn → RPn defined in Example2.13(f) is a smooth covering map.

Since Sn is compact, it is clear that q is proper. For each i = 1, . . . , n+ 1, let

S+i = (x1, . . . , xn+1) ∈ Sn : xi > 0 ,S−i = (x1, . . . , xn+1) ∈ Sn : xi < 0 .

Then q|S+i

and q|S−i are diffeomorphisms for i = 1, . . . , n + 1, which shows that q is a

local diffeomorphism. By Proposition 4.46, q is a smooth covering map.

Example 61. [Problem 4-13] Define a map F : S2 → R4 by F (x, y, z) = (x2 −y2, xy, xz, yz). Using the smooth covering map of Example 2.13(f) and Problem 4-10,show that F descends to a smooth embedding of RP2 into R4.

Let q : S2 → RP2 be the map of Example 60. Since F (−x,−y,−z) = F (x, y, z), the

map F descends to a unique smooth map F : RP2 → R4. It is easy to check that F−1

is smooth and given by

F−1(t, u, v, w) = q

(√uv

w,

√uw

v,

√vw

u

).

Note that we take uv/w = 0 if w = 0, since w = yz = 0 implies that either u = 0 or

v = 0. This is true for the other coordinates. Therefore F is a homeomorphism ontoits image. It remains to show that F is a smooth immersion. Let

ϕ : RP2 \

[x, y, z] ∈ RP2 : z 6= 0→ R2

[x, y, z] 7→(xz,y

z

)be coordinate map. Then

(F ϕ−1)(x, y) = F [x, y, 1]

= (x2 − y2, xy, x, y)

so that

D(F ϕ−1)(x, y) =

2x −2yy x1 00 1

,which clearly has full rank. We can perform similar computations for the other coordi-

nate maps, showing that dF[x,y,z] is always surjective.

21

Chapter 5. Submanifolds

Example 62. [Exercise 5.10] Show that spherical coordinates (Example C.38) form aslice chart for S2 in R3 on any open subset where they are defined.

Spherical coordinates are given by

(ρ, ϕ, θ) 7→ (ρ sinϕ cos θ, ρ sinϕ sin θ, ρ cosϕ).

For the image to be a subset of S2, we must have ρ = 1.

Theorem 63. [Exercise 5.24] Suppose M is a smooth manifold with or without bound-ary, S ⊆ M is an immersed k-submanifold, ι : S → M is the inclusion map, and Uis an open subset of Rk. A map X : U → M is a smooth local parametrization of Sif and only if there is a smooth coordinate chart (V, ϕ) for S such that X = ι ϕ−1.Therefore, every point of S is in the image of some local parametrization.

Proof. If X is a smooth local parametrization of S then X−1 : X(U) → U is a diffeo-morphism, so (X(U), X−1) is a smooth chart such that X = ι (X−1)−1. Conversely,suppose that (V, ϕ) is a smooth chart such that X = ι ϕ−1. Then X(ϕ(V )) = V isopen in S and X is a diffeomorphism onto V , which shows that X is a smooth localparametrization.

Theorem 64. [Exercise 5.36] Suppose M is a smooth manifold with or without bound-ary, S ⊆ M is an immersed or embedded submanifold, and p ∈ S. A vector v ∈ TpMis in TpS if and only if there is a smooth curve γ : J →M whose image is contained inS, and which is also smooth as a map into S, such that 0 ∈ J , γ(0) = p, and γ′(0) = v.

Proof. Let ι : S →M be the inclusion map. If v ∈ TpS then by Proposition 3.23 thereis a smooth curve γ0 : J → S such that 0 ∈ J , γ0(0) = p and γ′0(0) = v. Thereforeγ = ι γ0 is the desired smooth curve. Conversely, if there is a smooth curve γ withthe specified properties then γ0 : J → S given by t 7→ γ(t) is smooth. Thereforeγ′0(0) ∈ TpS and v = γ′(0) = dιp(γ

′0(0)), so v is in TpS (as a subspace of TpM).

Theorem 65. [Exercise 5.40] Suppose S ⊆M is a level set of a smooth map Φ : M →N with constant rank. Then TpS = ker dΦp for each p ∈ S.

Proof. Follows from Theorem 5.12 and Proposition 5.38.

Theorem 66. [Exercise 5.42] Suppose M is a smooth n-dimensional manifold withboundary, p ∈ ∂M , and (xi) are any smooth boundary coordinates defined on a neigh-borhood of p. The inward-pointing vectors in TpM are precisely those with positivexn-component, the outward-pointing ones are those with negative xn-component, andthe ones tangent to ∂M are those with zero xn-component. Thus, TpM is the disjointunion of Tp∂M , the set of inward-pointing vectors, and the set of outward-pointingvectors, and v ∈ TpM is inward-pointing if and only if −v is outward-pointing.

22

Proof. As in Proposition 3.23.

Theorem 67. [Exercise 5.44] Suppose M is a smooth manifold with boundary, f is aboundary defining function, and p ∈ ∂M . A vector v ∈ TpM is inward-pointing if andonly if vf > 0, outward-pointing if and only if vf < 0, and tangent to ∂M if and onlyif vf = 0.

Proof. Let (x1, . . . , xn) be smooth coordinates defined on a neighborhood of p. Since f isa boundary defining function, (∂f/∂xi)(p) = 0 for i = 1, . . . , n−1 and (∂f/∂xn)(p) > 0.Write v =

∑ni=1 v

i(∂/∂xi|p) so that

vf =n∑i=1

vi∂f

∂xi(p) = vn

∂f

∂xn(p);

then the result follows from Theorem 66.

Example 68. [Problem 5-1] Consider the map Φ : R4 → R2 defined by

Φ(x, y, s, t) = (x2 + y, x2 + y2 + s2 + t2 + y).

Show that (0, 1) is a regular value of Φ, and that the level set S = Φ−1((0, 1)) isdiffeomorphic to S2.

We compute

DΦ(x, y, s, t) =

[2x 1 0 02x 2y + 1 2s 2t

].

This matrix clearly has full rank when s 6= 0 or t 6= 0, so suppose that s = t = 0. Thedeterminant of the submatrix formed by the first two columns is

2x(2y + 1)− 2x = 4xy,

which is zero if and only if x = 0 or y = 0. In either case, we cannot have Φ(x, y, s, t) =(0, 1). This shows that dΦp is surjective for all p ∈ S. Define ϕ : R4 → R4 by

ϕ(x, y, s, t) = (x, x2 + y, s, t);

then ϕ is a diffeomorphism and Φ = (Φ ϕ−1)(u, v, s, t) = (v, v + (v − u2)2 + s2 + t2).It is easy to see that

ϕ(S) =

(u, 0, s, t) ∈ R4 : u4 + s2 + t2 = 1.

This set can be considered as an embedded submanifold of R4 diffeomorphic to S2, suchthat ϕ|S is a diffeomorphism. Therefore S is diffeomorphic to ϕ(S).

Theorem 69. [Problem 5-2] If M is a smooth n-manifold with boundary, then with thesubspace topology, ∂M is a topological (n−1)-dimensional manifold (without boundary),and has a smooth structure such that it is a properly embedded submanifold of M .

23

Proof. Let x ∈ ∂M and choose a smooth boundary chart (U,ϕ) containing x. Then

ϕ(∂M ∩ U) =

(x1, . . . , xn) ∈ ϕ(U) : xn = 0,

which shows that ∂M satisfies the local (n− 1)-slice condition. The result follows fromTheorem 5.8 and Proposition 5.5.

Theorem 70. [Problem 5-3] Suppose M is a smooth manifold with or without boundary,and S ⊆ M is an immersed submanifold. If any of the following holds, then S isembedded.

(1) S has codimension 0 in M .(2) The inclusion map S ⊆M is proper.(3) S is compact.

Proof. If (1) holds then Theorem 4.5 shows that the inclusion map ι : S → M is anopen map. The result follows from Proposition 4.22.

Example 71. [Problem 5-4] Show that the image of the curve β : (−π, π) → R2 ofExample 4.19 given by

β(t) = (sin 2t, sin t)

is not an embedded submanifold of R2.

It is well-known that the image of β with the subset topology is not a topologicalmanifold at all. (This can be seen by considering the point (0, 0), around which thereis no coordinate map into R.)

Example 72. [Problem 5-5] Let γ : R→ T2 be the curve of Example 4.20. Show thatγ(R) is not an embedded submanifold of the torus.

Again, γ(R) with the subset topology is not a topological manifold because it is notlocally Euclidean.

Theorem 73. [Problem 5-6] Suppose M ⊆ Rn is an embedded m-dimensional subman-ifold, and let UM ⊆ TRn be the set of all unit tangent vectors to M :

UM = (x, v) ∈ TRn : x ∈M, v ∈ TxM, |v| = 1 .

It is called the unit tangent bundle of M . Prove that UM is an embedded (2m−1)-dimensional submanifold of TRn ≈ Rn × Rn.

Proof. Let (x, v) ∈ UM . Since M is an embedded submanifold of Rn, we can choose asmooth chart (U,ϕ) for Rn containing x such that

ϕ(M ∩ U) =

(x1, . . . , xn) ∈ ϕ(U) : xm+1 = · · · = xn = 0.

24

Similarly, Sm−1 is an embedded submanifold of Rm, so we can choose a smooth chart(V, ψ) for Rm containing v such that

ψ(Sm−1 ∩ V ) =

(x1, . . . , xm) ∈ ψ(V ) : xm = 0.

Write ϕ = (x1, . . . , xn) and U = ϕ−1(U). By definition, the map ϕ : U → R2n given by

vi∂

∂xi

∣∣∣∣p

7→ (x1(p), . . . , xn(p), v1, . . . , vn)

is a coordinate map for TRn. By shrinking U , we can assume that V ⊆ π(ϕ(U)), whereπ : R2n → Rm is the projection onto the coordinates n+ 1, . . . , n+m. Define

θ(x1, . . . , xn, v1, . . . , vn) = (x1, . . . , xn, ψ(v1, . . . , vm), vm+1, . . . , vn);

then θ is a diffeomorphism onto its image, and θ ϕ : U → R2n is still a coordinatemap for TRn. Furthermore,

(θ ϕ)(UM ∩ U)

=

(x1, . . . , xn, v1, . . . , vn) ∈ (θ ϕ)(U) : xm+1 = · · · = xn = vm = · · · = vn = 0,

so UM satisfies the local (2m−1)-slice condition. By Theorem 5.8, UM is an embedded(2m− 1)-dimensional submanifold of TRn.

Example 74. [Problem 5-7] Let F : R2 → R be defined by F (x, y) = x3 + xy + y3.Which level sets of F are embedded submanifolds of R2? For each level set, prove eitherthat it is or that it is not an embedded submanifold.

We computeDF (x, y) =

[3x2 + y x+ 3y2

],

which has full rank unless 3x2 + y = 0 and x+ 3y2 = 0. In this case, we have

y = −3x2 = −3(−3y2)2 = −27y4

and0 = y + 27y4 = y(3y + 1)(9y2 − 3y + 1),

so y = 0 or y = −13. By symmetry, we have x = 0 or x = −1

3. From this, it is clear

that DF (x, y) has full rank if and only if (x, y) 6= (0, 0) and (x, y) 6= (−13,−1

3). Note

that F (0, 0) = 0 and F (−13,−1

3) = 1

27. We first examine the level set S1 = F−1(0).

The point p = (0, 0) is a saddle point of F , so S1 is a self-intersecting curve at p. Next,we examine the level set S2 = F−1(

127

). Since F has a strict local maximum at

p = (−13,−1

3), the point p is isolated in S2. In both cases, the level sets fail to be locally

Euclidean. Therefore, we can conclude that F−1(c) is an embedded 1-dimensionalsubmanifold of R2 if and only if c 6= 0 and c 6= 1

27.

Theorem 75. Any closed ball in Rn is an n-dimensional manifold with boundary.

25

Proof. Let Bn = B1(0) be the closed unit ball in Rn, let N = (0, . . . , 0, 1) be the “northpole”, and let T = 0n−1 × [0, 1] be the vertical line connecting 0 and N . Defineσ : Bn \ T → Hn given by

σ(x1, . . . , xn) =

(x1

‖x‖ − xn, . . . ,

xn−1

‖x‖ − xn, ‖x‖−1 − 1

),

with its inverse given by

σ−1(u1, . . . , un) =1

(un + 1)(‖u‖2 + 1)(2u1, . . . , 2un−1, ‖u‖2 − 1)

where u = (u1, . . . , un−1). Since both maps are continuous, this proves that Bn \ T ishomeomorphic to Hn. Furthermore, we can repeat the same argument by taking thevertical line T = 0n−1 × [−1, 0] instead, giving us Euclidean neighborhoods of everypoint in Bn except for 0. But the identity map on the open unit ball Bn is a suitablecoordinate chart around 0, so this proves that Bn is an n-manifold with boundary.

Theorem 76. [Problem 5-8] Suppose M is a smooth n-manifold and B ⊆M is a regularcoordinate ball. Then M \ B is a smooth manifold with boundary, whose boundary isdiffeomorphic to Sn−1.

Proof. We have coordinate balls around every point of M \B, so it remains to show thatwe have coordinate half-balls around each point of ∂B. Since B is a regular coordinateball, this reduces to showing that Br(0) \ Bs(0) is a smooth manifold with boundarywhenever s < r. But this follows easily from the stereographic projection described inTheorem 75.

Theorem 77. [Problem 5-9] Let S ⊆ R2 be the boundary of the square of side 2 centeredat the origin (see Theorem 36). Then S does not have a topology and smooth structurein which it is an immersed submanifold of R2.

Proof. Suppose S has a topology and smooth structure for which the inclusion mapι : S → R2 is a smooth immersion. By Theorem 4.25, there is a neighborhood Uof (1, 1) such that ι|U is a smooth embedding. We can then apply the argument ofTheorem 36 to derive a contradiction.

Example 78. [Problem 5-10] For each a ∈ R, let Ma be the subset of R2 defined by

Ma =

(x, y) : y2 = x(x− 1)(x− a).

For which values of a is Ma an embedded submanifold of R2? For which values can Ma

be given a topology and smooth structure making it into an immersed submanifold?Let F (x, y) = y2 − x(x− 1)(x− a); then

DF (x, y) =[−(3x2 − (2a+ 2)x+ a) 2y

].

26

Therefore 0 is a regular value of F unless there is a point (x, y) such that y = 0,3x2 − (2a + 2)x + a = 0 and F (x, y) = −x(x − 1)(x − a) = 0. We have the followingcases to consider:

a = 0 and (x, y) = (0, 0),

a = 1 and (x, y) = (1, 0).

When a = 0 the point (0, 0) is a local minimum of F , so (0, 0) is an isolated point ofM0. Therefore M0 cannot be an embedded or immersed submanifold. When a = 1 thecurve M1 is self-intersecting at (1, 0). By giving M1 an appropriate topology in whichit is disconnected, we can make M1 an immersed submanifold of R2.

Theorem 79. [Problem 5-12] Suppose E and M are smooth manifolds with boundary,and π : E → M is a smooth covering map. The restriction of π to each connectedcomponent of ∂E is a smooth covering map onto a component of ∂M .

Proof. Let F be a connected component of ∂E. Let x ∈ π(F ), let U be an evenly

covered neighborhood of x, choose e so that x = π(e), and let U be the component

of π−1(U) containing e. Then π|U : U → U is a diffeomorphism. It is clear that

π(U ∩F ) = U ∩∂M . Since U ∩F is an embedded submanifold of U and π(U ∩F ) is anembedded submanifold of U by Theorem 5.11, π|U∩F is a diffeomorphism. This showsthat U ∩∂M is an evenly covered neighborhood of x, and that the image of π|F is open.Let x /∈ π(F ), let U be an evenly covered neighborhood of x, and let V = U ∩ ∂M . Byshrinking V if necessary, we may assume that V is a connected neighborhood of x in

U ∩ ∂M . Suppose there is a f ∈ F such that y = π(f) ∈ V . Let U be the component

of π−1(U) containing f and choose e ∈ U so that x = π(e). Let γ be a path from x

to y in V . Then (π|U)−1 γ is a path from e to f in U ∩ ∂E, so e ∈ F since f ∈ Fand F is a connected component of ∂E. This contradicts the fact that x /∈ π(F ), soV ⊆ ∂M \ π(F ). This shows that π(F ) is both open and closed in ∂M , so it is acomponent of ∂M .

Theorem 80. [Problem 5-14] Suppose M is a smooth manifold and S ⊆ M is animmersed submanifold. For the given topology on S, there is only one smooth structuremaking S into an immersed submanifold.

Proof. Let ι : S →M be the inclusion map, which is a smooth immersion. Let S ′ denoteS with some other smooth structure such that ι′ : S ′ → M is a smooth immersion.Since S and S ′ have the same topology, the maps ι : S → S ′ and ι′ : S ′ → S areboth continuous, and Theorem 5.29 shows that the maps are inverses of each other.Therefore S is diffeomorphic to S ′.

27

Theorem 81. [Problem 5-16] If M is a smooth manifold and S ⊆ M is a weaklyembedded submanifold, then S has only one topology and smooth structure with respectto which it is an immersed submanifold.

Proof. Almost identical to Theorem 80.

Theorem 82. [Problem 5-17] Suppose M is a smooth manifold and S ⊆M is a smoothsubmanifold.

(1) S is embedded only if every f ∈ C∞(S) has a smooth extension to a neighborhoodof S in M .

(2) S is properly embedded only if every f ∈ C∞(S) has a smooth extension to allof M .

Proof. First suppose that S is embedded. Let p ∈ S and choose a smooth slice chart(Up, ϕp) for S in M containing p. Then S∩Up is closed in Up, so Proposition 2.25 showsthat there is smooth function fp : Up → R such that fp|S∩Up = f |S∩Up . Let ψpp∈S be

a partition of unity subordinate to Upp∈S; then

f(x) =∑p∈S

ψp(x)fp(x)

is a smooth function defined on U =⋃p∈S Up such that f |S = f , taking fp to be zero

on U \ supp fp. For (2), Proposition 5.5 shows that S is closed in M . Therefore we have

a partition of unity ψp ∪ ψ0 for M subordinate to Up ∪ M \ S, allowing f tobe defined on all of M .

Theorem 83. [Problem 5-19] Suppose S ⊆M is an embedded submanifold and γ : J →M is a smooth curve whose image happens to lie in S. Then γ′(t) is in the subspaceTγ(t)S of Tγ(t)M for all t ∈ J . This need not be true if S is not embedded.

Proof. Corollary 5.30 shows that γ is smooth as a map into S. Denote this map byγ0 : J → S and let ι : S → M . Then γ = ι γ0, so γ′(t) = dιγ(t)(γ

′0(t)) by Proposition

3.24 and therefore γ′(t) ∈ Tγ(t)S ⊆ Tγ(t)M . This need not hold if S is merely animmersed submanifold. For example, consider a curve γ that crosses the point of self-intersection in the figure-eight curve β of Example 4.19, such that γ is not continuousas a map into the image of β.

Theorem 84. [Problem 5-21] Suppose M is a smooth manifold and f ∈ C∞(M).

(1) For each regular value b of f , the sublevel set f−1 ((−∞, b]) is a regular domainin M .

(2) If a and b are two regular values of f with a < b, then f−1 ([a, b]) is a regulardomain in M .

28

Proof. Let m be the dimension of M . It is clear that f−1 ((−∞, b)) satisfies the localm-slice condition. Furthermore, there is a boundary slice chart around each pointof f−1(b). We can see this by applying the argument in Corollary 5.14, with thefollowing modification to Theorem 5.12: for each p ∈ f−1(b), the rank theorem showsthat there are smooth charts (U,ϕ) centered at p and (V, ψ) containing b for which the

coordinate representation f of f has the form

f(x1, . . . , xm) = x1.

By modifying f and ψ appropriately, U becomes a boundary slice chart for f−1 ((−∞, b])in M . Theorem 5.51 then shows that f−1 ((−∞, b]) is an embedded submanifold withboundary in M , and it is properly embedded since it is closed in M . Part (2) issimilar.

Theorem 85. [Problem 5-22] If M is a smooth manifold and D ⊆ M is a regulardomain, then there exists a defining function for D. If D is compact, then f can betaken to be a smooth exhaustion function for M .

Proof. Let U = (Up, ϕp)p∈D be a collection of interior or boundary slice charts for D

in M . Then U ∪ M \D covers M , and we can repeat the argument of Proposition5.43 by constructing functions fpp∈D ∪ f0 as follows: set fp = −1 when p ∈ IntD,

set fp(x1, . . . , xn) = −xn when p ∈ ∂D, and set f0 = 1 for f0 : M \ D → R. If D is

compact then we can modify f by using the argument of Proposition 2.28, taking acountable open cover of M \D by precompact open subsets.

Chapter 6. Sard’s Theorem

Theorem 86. [Exercise 6.7] Let M be a smooth manifold with or without boundary.Any countable union of sets of measure zero in M has measure zero.

Proof. Let A =⋃∞n=1An where each An has measure zero. Let (U,ϕ) be a smooth

chart; we want to show that

A ∩ U =∞⋃n=1

An ∩ U

has measure zero. But this follows from the fact that

ϕ

(∞⋃n=1

An ∩ U

)=∞⋃n=1

ϕ(An ∩ U)

has measure zero (in Rn) since each ϕ(An ∩ U) has measure zero.

29

Theorem 87. [Problem 6-1] Suppose M and N are smooth manifolds with or withoutboundary, and F : M → N is a smooth map. If dimM < dimN , then F (M) hasmeasure zero in N .

Proof. Let n = dimN . Define F : M × Rk → N by (x, y) 7→ F (x), where k =dimN − dimM . This map is smooth because it can be written as the compositionπ (F × IdRk), where π : N × Rk → N is the canonical projection. It suffices to show

that the image of F has measure zero in N . Let (U1×U2, ϕ1×ϕ2) and (V, ψ) be smoothcharts for M×Rk and N respectively such that U1 ⊆M and U2 ⊆ Rk. Let U = U1×U2

and ϕ = ϕ1 × ϕ2. Then

ψ(F (U) ∩ V ) = ψ(F (U ∩ F−1(V )))

= (ψ F ϕ−1)(ϕ(U ∩ F−1(V )))

⊆ (ψ F ϕ−1)(ϕ(U) ∩ (ϕ1(F−1(V ))× 0)),

which has measure zero in Rn by Proposition 6.5. Therefore F (U) has measure zero.

But F (M × Rk) can be written as a union of countably many sets of this form, so

F (M × Rk) = F (M) has measure zero in N .

Theorem 88. [Problem 6-2] Every smooth n-manifold (without boundary) admits asmooth immersion into R2n.

Proof. Let M be a smooth n-manifold; by Theorem 6.15, we can assume that M is anembedded submanifold of R2n+1. Let UM ⊆ TR2n+1 be the unit tangent bundle of M(see Theorem 73) and define G : UM → RP2n by (x, v) 7→ [v]. This map is smooth

because it is the composition UM → R2n+1 \ 0 π−→ RP2n. Since dimUM < dimRP2n,Sard’s theorem shows that the image of G has measure zero in RP2n, and since π(R2n)has measure zero in RP2n, there is a v ∈ R2n+1 \ R2n such that [v] is not in the imageof G. Let πv : R2n+1 → R2n be the projection with kernel Rv; then πv|M is smooth.Furthermore, if p ∈M then v /∈ TpM , so d(πv|M)p is injective.

Theorem 89. [Problem 6-3] Let M be a smooth manifold, let B ⊆ M be a closedsubset, and let δ : M → R be a positive continuous function. There is a smooth

function δ : M → R that is zero on B, positive on M \ B, and satisfies δ(x) < δ(x)everywhere.

Proof. By Theorem 2.29, there is a smooth nonnegative function f : M → R such thatf−1(0) = B. By replacing f with f/(f + 1), we can assume that f = 0 on B and0 < f ≤ 1 on M \ B. By Corollary 6.22, there is a smooth e : M → R such that

0 < e(x) < δ(x) for all x ∈M . Then δ = fe is the desired function.

30

Example 90. [Problem 6-7] By considering the map F : R→ H2 given by F (t) = (t, |t|)and the subset A = [0,∞) ⊆ R, show that the conclusions of Theorem 6.26 andCorollary 6.27 can be false when M has nonempty boundary.

F is smooth on the closed subset [0,∞), but there is no smooth map G : R→ H2 suchthat G|A = F |A.

Theorem 91. [Problem 6-8] Every proper continuous map between smooth manifoldsis homotopic to a proper smooth map.

Proof. Since the retraction r : U →M is proper, it suffices to show that if F is proper

in Theorem 6.21, then F is also proper. Let E ⊆ Rk be compact. Since δ is continuous,it attains some maximum δ(x0) on E. Let B be a closed ball containing E such thatfor every x ∈ E, the closed ball of radius δ(x) around x is contained in B. Since

|F (x) − F (x)| < δ(x), it is clear that F−1(E) is a closed subset of F−1(B). But F is

proper, so F−1(B) is compact and therefore F−1(E) is compact.

Example 92. [Problem 6-9] Let F : R2 → R3 be the map

F (x, y) = (ey cosx, ey sinx, e−y).

For which positive numbers r is F transverse to the sphere Sr(0) ⊆ R3? For whichpositive numbers r is F−1 (Sr(0)) an embedded submanifold of R2?

We compute

DF (x, y) =

−ey sinx ey cosxey cosx ey sinx

0 −e−y

.We have ‖F (x, y)‖ = r if and only if

r2 = e2y cos2 x+ e2y sin2 x+ e−2y = e2y + e−2y,

i.e.

y =1

2log

(1

2(r2 ±

√r4 − 4)

).

Thus F−1 (Sr(0)) is always an embedded submanifold of R2, being either empty (r <√2), one line (r =

√2) or two lines in R2 (r >

√2). However, F is transverse to Sr(0)

if and only if r >√

2. Let us examine the case r =√

2: we have y = 0, so the image ofdF(x,y) is spanned by the columns of

DF (x, 0) =

− sinx cosxcosx sinx

0 −1

.

31

The corresponding point in S√2(0) is v(x) = (cos x, sinx, 1), so the tangent space con-sists of all vectors orthogonal to v(x). But v(x) is orthogonal to the columns ofDF (x, 0),so dF(x,y) + Tv(x)S√2(0) is a proper subset of Tv(x)R3.

Theorem 93. [Problem 6-15] Suppose M and N are smooth manifolds and S ⊆M×Nis an immersed submanifold. Let πM and πN denote the projections from M ×N ontoM and N , respectively. The following are equivalent:

(1) S is the graph of a smooth map f : M → N .(2) πM |S is a diffeomorphism from S onto M .(3) For each p ∈ M , the submanifolds S and p × N intersect transversely in

exactly one point.

If these conditions hold, then S is the graph of the map f : M → N defined by f =πN (πM |S)−1.

Proof. Let m = dimM and n = dimN . Suppose S = (x, f(x)) : x ∈M for somesmooth map f : M → N . Then S is the image of the map g : M → S given byx 7→ (x, f(x)), so πM |S is smooth with inverse g. Conversely, if πM |S is a diffeomorphismthen we can take f = πN (πM |S)−1 in (1). This proves that (1) ⇔ (2). If (1) holdsthen d(πM)g(p) dgp is always nonzero, so TpS + Tp(p ×N) = Tp(M ×N). It is alsoclear that S ∩ (p × N) consists of exactly one point. If (3) holds then πM |S is abijection. For every p ∈ M , the submanifolds S and p × N intersect transversely.Since dim(p × N) = n < dim(M × N), we must have dim d(πM |S)p(TpS) ≥ m.Therefore d(πM |S)p is invertible for every p ∈ M , and πM |S is a diffeomorphism. Thisproves that (1)⇔ (3).

Chapter 7. Lie Groups

Theorem 94. [Exercise 7.2] If G is a smooth manifold with a group structure suchthat the map t : G×G→ G given by (g, h) 7→ gh−1 is smooth, then G is a Lie group.

Proof. The inversion map i(g) = t(e, g) is smooth, and so is the multiplication mapm(g, h) = t(g, i(h)) = t(g, t(e, h)).

Theorem 95. [Problem 7-1] For any Lie group G, the multiplication map m : G×G→G is a smooth submersion.

Proof. For each g ∈ G, the map σg : G → G × G given by x 7→ (g, g−1x) is a smoothlocal section of m since m(σg(x)) = gg−1x = x. Every point (g1, g2) ∈ G×G is in theimage of σg1 , so Theorem 4.26 shows that m is a smooth submersion.

Theorem 96. [Problem 7-2] Let G be a Lie group.

32

(1) Let m : G × G → G denote the multiplication map. Using Proposition 3.14 toidentify T(e,e)(G×G) with TeG⊕TeG, the differential dm(e,e) : TeG⊕TeG→ TeGis given by

dm(e,e)(X, Y ) = X + Y.

(2) Let i : G→ G denote the inversion map. Then die : TeG→ TeG is given by

die(X) = −X.

Proof. We have

dm(e,e)(X, Y ) = dm(e,e)(X, 0) + dm(e,e)(0, Y )

= d(m(1))e(X) + d(m(2))e(Y ),

where m(1) : G→ G is given by x 7→ m(x, e) and m(2) : G→ G is given by y 7→ m(e, y).But m(1) = m(2) = IdG, so dm(e,e)(X, Y ) = X + Y . This proves (1). Let n = m p swhere s : G → G × G is given by x 7→ (x, x) and p : G × G → G × G is given by(x, y) 7→ (x, i(y)); then n is constant, so

0 = dne(X)

= dm(e,e)(dp(e,e)(dse(X)))

= dm(e,e)(dp(e,e)(X,X))

= dm(e,e)(X, die(X))

= X + die(X).

Therefore die(X) = −X.

Theorem 97. [Problem 7-3] If G is a smooth manifold with a group structure suchthat the multiplication map m : G×G→ G is smooth, then G is a Lie group.

Proof. It suffices to show that the inversion map i : G → G is smooth. Define F :G×G→ G×G by (g, h) 7→ (g, gh), which is smooth and bijective. We have

dF(e,e)(X, Y ) = (X,X + Y ),

so dF(e,e) is an isomorphism. Write Lg and Rg for the translation maps. If (x, y) ∈ G×Gthen it is easy to check that

F = ((Lxy Ry−1)× IdG) F ((Ly−1x−1 Ry)× Ly−1),

so dF(x,y) = ϕ dF(e,e) ψ where ϕ, ψ are isomorphisms. Therefore dF(x,y) is an isomor-phism and F is a bijective local diffeomorphism, i.e. a diffeomorphism. Since i(g) isthe second component of F−1(g, e), the inversion map i is smooth.

Theorem 98. [Problem 7-4] Let det : GL(n,R)→ R denote the determinant function.Then

d(det)X(B) = (detX) tr(X−1B).

33

Proof. Let A ∈ M(n,R) write Ai,j for the (i, j) entry of A. Then

d

dt

∣∣∣∣t=0

det(In + tA) =d

dt

∣∣∣∣t=0

(1 + tr(A)t+ · · · )

= tr(A).

If X ∈ GL(n,R) and B ∈ TX GL(n,R) we have

det(X + tB) = det(X) det(In + tX−1B),

so

d(det)X(B) =d

dt

∣∣∣∣t=0

det(X + tB)

= det(X)d

dt

∣∣∣∣t=0

det(In + tX−1B)

= det(X) tr(X−1B).

Theorem 99. [Problem 7-5] For any connected Lie group G, the universal covering

group is unique in the following sense: if G and G′ are simply connected Lie groups

that admit smooth covering maps π : G → G and π′ : G′ → G that are also Lie group

homomorphisms, then there exists a Lie group isomorphism Φ : G → G′ such thatπ′ Φ = π.

Proof. Let e be the identity in G and let e′ be the identity in G′. By Theorem 50 there

is a diffeomorphism Φ : G→ G′ such that π′ Φ = π, so it remains to show that Φ is agroup homomorphism. By replacing Φ with Φ(e)−1Φ = LΦ(e)−1 Φ, we can assume thatΦ(e) = e′. (Since π′(Φ(e)−1Φ(g)) = π′(Φ(e))−1π(g) = π(g), we still have π′ Φ = π.) If

g, h ∈ G then

π′(Φ(gh)) = π(g)π(h) = π′(Φ(g)Φ(h)),

so (g, h) 7→ Φ(gh) and (g, h) 7→ Φ(g)Φ(h) are both lifts of (g, h) 7→ π(gh). Since themaps all agree at the point (e, e), we have Φ(gh) = Φ(g)Φ(h).

Theorem 100. [Problem 7-6] Suppose G is a Lie group and U is any neighborhoodof the identity. There exists a neighborhood V of the identity such that V ⊆ U andgh−1 ∈ U whenever g, h ∈ V .

Proof. Define f(g, h) = gh−1 and let W = f−1(U). Since (e, e) ∈ W , there are neigh-borhoods W1,W2 of e such that (e, e) ∈ W1 ×W2 ⊆ W . Then V = W1 ∩W2 is thedesired neighborhood of the identity.

34

Theorem 101. [Problem 7-7] Let G be a Lie group and let G0 be its identity component.Then G0 is a normal subgroup of G, and is the only connected open subgroup. Everyconnected component of G is diffeomorphic to G0.

Proof. The subgroup H generated by G0 is a connected open subgroup of G, so H ⊆ G0

since G0 is a connected component. Therefore H = G0. For each g ∈ G, define theconjugation map Cg : G→ G by h 7→ ghg−1. This map is a Lie group isomorphism, soCg(G0) is a connected open subgroup of G and again we have Cg(G0) = G0. Every leftcoset gG0 is the image of G0 under the diffeomorphism Lg, so gG0 is a connected opensubset of G. But G is the union of all such cosets, so the connected components of Gare precisely the cosets of G0.

Theorem 102. [Problem 7-8] Suppose a connected topological group G acts continu-ously on a discrete space K. Then the action is trivial.

Proof. Let θ : G ×K → K be the group action. For any x ∈ K the set θ(G × x) isconnected and therefore consists of one point. But θ(e, x) = x, so θ(g, x) = x for everyg ∈ G.

Theorem 103. [Problem 7-9,7-10] The formula

A · [x] = [Ax]

defines a smooth, transitive left action of GL(n + 1,R) on RPn. The same formuladefines a smooth, transitive left action of GL(n+ 1,C) on CPn.

Proof. The smooth map θ : GL(n + 1,R)× Rn+1 \ 0 → RPn given by (A, x) 7→ [Ax]is constant on each line in Rn+1 since A is linear, so it descends to a smooth mapθ : GL(n + 1,R) × RPn → RPn given by (A, [x]) 7→ [Ax]. It is easy to see that θ is agroup action. Furthermore, for any two points v, w ∈ Rn+1 there is an invertible linearmap that takes v to w, so θ is transitive.

Example 104. [Problem 7-11] Considering S2n+1 as the unit sphere in Cn+1, define anaction of S1 on S2n+1, called the Hopf action, by

z · (w1, . . . , wn+1) = (zw1, . . . , zwn+1).

This action is smooth and its orbits are disjoint unit circles in Cn+1 whose union isS2n+1.

Theorem 105. [Problem 7-12] Every Lie group homomorphism has constant rank.

Proof. If F : G → H is a Lie group homomorphism, then we can apply Theorem 7.25with the left translation action on G and the action g · h = F (g)h on H.

35

Theorem 106. [Problem 7-13,7-14] For each n ≥ 1, U(n) is a properly embeddedn2-dimensional Lie subgroup of GL(n,C) and SU(n) is a properly embedded (n2 − 1)-dimensional Lie subgroup of U(n).

Proof. We can use the argument in Example 7.27, replacing the transpose operatorwith the conjugate transpose operator.

Theorem 107. [Problem 7-15] SO(2), U(1), and S1 are all isomorphic as Lie groups.

Proof. U(1) is obviously identical to S1. We also have a Lie group isomorphism f :S1 → SO(2) given by

f(z) =

[Re(z) −Im(z)Im(z) Re(z)

]=

[cos θ − sin θsin θ cos θ

],

where θ = arg(z).

Theorem 108. [Problem 7-16] SU(2) is diffeomorphic to S3.

Proof. Identifying S3 with the subspace

(z, w) ∈ C2 : |z|2 + |w|2 = 1

, we have a dif-feomorphism

(z, w) 7→[z −ww z

].

Example 109. [Problem 7-17] Determine which of the following Lie groups are com-pact:

GL(n,R), SL(n,R), GL(n,C), SL(n,C), U(n), SU(n).

Clearly GL(n,R) and GL(n,C) are not compact. SL(n,R) and SL(n,C) are not com-pact, since the operator norm ofr r−1

I

∈ SL(n,R)

is unbounded as r → ∞. However, U(n) is compact since every unitary matrix hasoperator norm 1, and SU(n) is compact because it is a properly embedded subgroup ofU(n).

Theorem 110. [Problem 7-18] Suppose G is a Lie group, and N,H ⊆ G are Liesubgroups such that N is normal, N ∩H = e, and NH = G. Then the map (n, h) 7→nh is a Lie group isomorphism between N oθ H and G, where θ : H × N → N is theaction by conjugation: θh(n) = hnh−1. Furthermore, N and H are closed in G.

36

Proof. Let ϕ be the map (n, h) 7→ nh. If nh = e then n = h−1 ∈ N ∩H, so n = h = e.Also, ϕ is surjective since NH = G. Therefore ϕ is a bijection. We have

ϕ((n, h)(n′, h′)) = ϕ(nhn′h−1, hh′)

= nhn′h−1hh′

= nhn′h′

= ϕ(n, h)ϕ(n′, h′),

so ϕ is an isomorphism. Since ϕ is smooth, it is a Lie group isomorphism.

Theorem 111. [Problem 7-19] Suppose G, N , and H are Lie groups. Then G is iso-morphic to a semidirect product NoH if and only if there are Lie group homomorphismsϕ : G→ H and ψ : H → G such that ϕ ψ = IdH and kerϕ ∼= N .

Proof. Suppose there is an isomorphism f : G → N oθ H. Let π : N oθ H → Hbe the projection (n, h) 7→ h and let ι : H → N oθ H be the injection h 7→ (e, h).Take ϕ = π f and ψ = ι; then ϕ ψ = IdH and kerϕ ∼= N . Conversely, suppose thehomomorphisms ϕ and ψ exist. Let N = kerϕ and H1 = ψ(H). Then N is normal in Gand N ∩H1 = e, since x = ψ(h) and x ∈ kerϕ implies that h = ϕ(ψ(h)) = ϕ(x) = e.For any g ∈ G we have ϕ(ψ(ϕ(g))) = ϕ(g), so g = nψ(ϕ(g)) for some n ∈ kerϕ.Therefore NH1 = G. By Theorem 110, G is a semidirect product.

Example 112. [Problem 7-20] Prove that the following Lie groups are isomorphic tosemidirect products as shown.

(1) O(n) ∼= SO(n) o O(1).(2) U(n) ∼= SU(n) o U(1).(3) GL(n,R) ∼= SL(n,R) oR∗.(4) GL(n,C) ∼= SL(n,C) oC∗.

For (1) to (4), define ϕ by M 7→ det(M) and ψ by

z 7→[z

In−1

]where In−1 is the (n− 1)× (n− 1) identity matrix, and apply Theorem 111.

Theorem 113. [Problem 7-23] Let H∗ be the Lie group of nonzero quaternions, andlet S ⊆ H∗ be the set of unit quaternions. Then S is a properly embedded Lie subgroupof H∗, isomorphic to SU(2).

Proof. The norm |·| : H∗ → R∗ is a smooth homomorphism, since

|(a, b)| =√

(a, b)(a,−b) =

√|a|2 + |b|2

37

and |pq| = |p| |q|. Therefore ker |·| = S is a properly embedded Lie subgroup of H∗.Furthermore, we have a Lie group isomorphism

(a, b) 7→[a −bb a

]into SU(2).

Chapter 8. Vector Fields

Theorem 114. [Exercise 8.29] For X, Y ∈ X(M) and f, g ∈ C∞(M), we have

[fX, gY ] = fg[X, Y ] + (fXg)Y − (gY f)X.

Proof. We compute

[fX, gY ]h = fX(gY h)− gY (fXh)

= gfXY h+ Y hfXg − fgY Xh−XhgY f= fg[X, Y ]h+ (fXg)Y h− (gY f)Xh.

Theorem 115. [Exercise 8.34] Let A : g → h be a Lie algebra homomorphism. ThenkerA and imA are Lie subalgebras.

Proof. It suffices to show that kerA and imA are closed under brackets. If X, Y ∈ kerAthen A[X, Y ] = [AX,AY ] = 0, so [X, Y ] ∈ kerA. Similarly, if X, Y ∈ imA thenX = AX ′ and Y = AY ′ for some X ′, Y ′ ∈ g, so A[X ′, Y ′] = [AX ′, AY ′] = [X, Y ] and[X, Y ] ∈ imA.

Theorem 116. [Exercise 8.35] Suppose g and h are finite-dimensional Lie algebrasand A : g → h is a linear map. Then A is a Lie algebra homomorphism if and only ifA[Ei, Ej] = [AEi, AEj] for some basis (E1, . . . , En) of g.

Proof. One direction is obvious. Suppose that (E1, . . . , En) is a basis of g andA[Ei, Ej] =[AEi, AEj]. Let X, Y ∈ g and write X =

∑ni=1 xiEi and Y =

∑ni=1 yiEi; then

A[X, Y ] = A

[n∑i=1

xiEi,

n∑j=1

yjEj

]

=n∑i=1

xi

n∑j=1

yjA [Ei, Ej]

=n∑i=1

xi

n∑j=1

yj [AEi, AEj]

38

=

[A

n∑i=1

xiEi, A

n∑j=1

yjEj

]= [AX,AY ].

Theorem 117. [Exercise 8.43] If V is any finite-dimensional real vector space, thecomposition of canonical isomorphisms

Lie(GL(V ))→ TId GL(V )→ gl(V )

yields a Lie algebra isomorphism between Lie(GL(V )) and gl(V ).

Proof. Choose a basis B for V and let ϕ : GL(V ) → GL(n,R) be the associated Liegroup isomorphism. It is also a Lie algebra isomorphism gl(V ) → gl(n,R) since itis linear. By Corollary 8.31 there is a Lie algebra isomorphism ρ : Lie(GL(V )) →Lie(GL(n,R)) induced by the diffeomorphism ϕ, and by Proposition 8.41 there is a Liealgebra isomorphism ψ : Lie(GL(n,R))→ gl(n,R). Then ϕ−1 ψ ρ is the desired Liealgebra isomorphism.

Theorem 118. [Problem 8-1] Let M be a smooth manifold with or without boundary,and let A ⊆ M be a closed subset. Suppose X is a smooth vector field along A. Given

any open subset U containing A, there exists a smooth global vector field X on M such

that X|A = X and supp X ⊆ U .

Proof. For each p ∈ A, choose a neighborhood Wp of p and a smooth vector field Xp

on Wp that agrees with X on Wp ∩ A. Replacing Wp by Wp ∩ U we may assume thatWp ⊆ U . The family of sets Wp : p ∈ A ∪ M \ A is an open cover of M . Letψp : p ∈ A ∪ ψ0 be a smooth partition of unity subordinate to this cover, withsuppψp ⊆ Wp and suppψ0 ⊆M \ A.

For each p ∈ A, the product ψpXp is smooth on Wp by Proposition 8.8, and has a smoothextension to all of M if we interpret it to be zero on M\suppψp. (The extended functionis smooth because the two definitions agree on the open subset Wp \ suppψp where they

overlap.) Thus we can define X : M → TM by

Xx =∑p∈A

ψp(x)Xp|x.

Because the collection of supports suppψp is locally finite, this sum actually has onlya finite number of nonzero terms in a neighborhood of any point of M , and therefore

39

defines a smooth function. If x ∈ A, then ψ0(x) = 0 and Xp|x = Xx for each p suchthat ψp(x) 6= 0, so

Xx =∑p∈A

ψp(x)Xx =

(ψ0(x) +

∑p∈A

ψp(x)

)Xx = Xx,

so X is indeed an extension of X. It follows from Lemma 1.13(b) that

supp X =⋃p∈A

suppψp =⋃p∈A

suppψp ⊆ U.

Theorem 119. [Problem 8-2] Let c be a real number, and let f : Rn \ 0 → Rbe a smooth function that is positively homogeneous of degree c, meaning thatf(λx) = λcf(x) for all λ > 0 and x ∈ Rn \ 0. Then V f = cf , where V is the Eulervector field defined in Example 8.3.

Proof. We want to compute

(V f)(x) = Vxf =n∑i=1

xi∂f

∂xi(x).

Differentiating λcf(x) = f(λx) with respect to λ on (0,∞), we have

cλc−1f(x) = Df(λx)(x) =n∑i=1

xi∂f

∂xi(λx).

Replacing x with λ−1x gives

cλc−1f(λ−1x) = λ−1

n∑i=1

xi∂f

∂xi(x)

⇒cλc−1λ−cf(x) = λ−1

n∑i=1

xi∂f

∂xi(x)

⇒cf(x) =n∑i=1

xi∂f

∂xi(x) = (V f)(x).

Theorem 120. [Problem 8-3] Let M be a nonempty positive-dimensional smooth man-ifold with or without boundary. Then X(M) is infinite-dimensional.

40

Proof. Let n = dimM . If n ≥ 2 then the result is clear from Proposition 8.7 sincethere are infinitely many lines in R2, i.e. RP1 is infinite. Suppose n = 1 and chooseany diffeomorphism ϕ : (−ε, 1 + ε) → U where ε > 0 and U is open in M . For eachm = 1, 2, . . . , define a vector field Xm along 1/k : k = 1, . . . ,m by setting

Xm|1/k =

0, k < m,ddt

∣∣1/k

, k = m.

For each m, we have an extension Xm defined on (−ε, 1 + ε) by Lemma 8.6. We will

show that Xm : m = 1, 2, . . . is linearly independent by induction. Since Xm is

nonzero, the set Xm0 is linearly independent for any m0. Now suppose

a1Xm1 + · · ·+ arXmr = 0

with m1 < · · · < mr. Then

a1Xm1|1/m1 + · · ·+ arXmr |1/m1 = a1Xm1|1/m1 = 0,

so a1 = 0 since Xm1 |1/m1 6= 0, and a2 = · · · = ar = 0 by induction. Therefore

ϕ∗Xm : m = 1, 2, . . . is a set of linearly independent vector fields defined on U . We

can extend these vector fields to M by restricting each ϕ∗Xm to the closed set ϕ([0, 1])and applying Lemma 8.6. This proves that X(M) is infinite-dimensional.

Theorem 121. [Problem 8-4] Let M be a smooth manifold with boundary. Thereexists a global smooth vector field on M whose restriction to ∂M is everywhere inward-pointing, and one whose restriction to ∂M is everywhere outward-pointing.

Proof. Let n = dimM . Let (Uα, ϕα) be a collection of smooth boundary charts whosedomains cover ∂M . For each α, let Xα be the nth coordinate vector field. Let ψαbe a partition of unity of U =

⋃α Uα subordinate to Uα; then X =

∑α ψαXα is a

smooth vector field defined on U . For any boundary defining function f : M → [0,∞)we have

Xpf =∑α

ψαXα|pf > 0

since each Xα|p is inward-pointing, so Xp is always inward-pointing. By restricting X to

∂M and applying Lemma 8.6, we have a global smooth vector field X whose restriction

to ∂M is everywhere inward-pointing. Also, the restriction of −X to ∂M is everywhereoutward-pointing.

Theorem 122. [Problem 8-5] Let M be a smooth n-manifold with or without boundary.

(1) If (X1, . . . , Xk) is a linearly independent k-tuple of smooth vector fields on anopen subset U ⊆ M , with 1 ≤ k < n, then for each p ∈ U there exist smooth

41

vector fields Xk+1, . . . , Xn in a neighborhood V of p such that (X1, . . . , Xn) is asmooth local frame for M on U ∩ V .

(2) If (v1, . . . , vk) is a linearly independent k-tuple of vectors in TpM for somep ∈ M , with 1 ≤ k ≤ n, then there exists a smooth local frame (Xi) on aneighborhood of p such that Xi|p = vi for i = 1, . . . , k.

(3) If (X1, . . . , Xn) is a linearly independent n-tuple of smooth vector fields along

a closed subset A ⊆ M , then there exists a smooth local frame (X1, . . . , Xn) on

some neighborhood of A such that Xi|A = Xi for i = 1, . . . , n.

Proof. See Theorem 162.

Theorem 123. [Problem 8-6] Let H be the algebra of quaternions and let S ⊆ H bethe group of unit quaternions.

(1) If p ∈ H is imaginary, then qp is tangent to S at each q ∈ S.(2) Define vector fields X1, X2, X3 on H by

X1|q = qi, X2|q = qj, X3|q = qk.

These vector fields restrict to a smooth left-invariant global frame on S.(3) Under the isomorphism (x1, x2, x3, x4)↔ x11 + x2i+ x3j + x4k between R4 and

H, these vector fields have the following coordinate representations:

X1 = −x2 ∂

∂x1+ x1 ∂

∂x2+ x4 ∂

∂x3− x3 ∂

∂x4,

X2 = −x3 ∂

∂x1− x4 ∂

∂x2+ x1 ∂

∂x3+ x2 ∂

∂x4,

X3 = −x4 ∂

∂x1+ x3 ∂

∂x2− x2 ∂

∂x3+ x1 ∂

∂x4.

Proof. S is a level set of the norm squared |·|2 : H∗ → (0,∞), which is given by

|(a+ bi, c+ di)|2 = a2 + b2 + c2 + d2

and has derivative [2a 2b 2c 2d

].

If q = (a, b, c, d) then by Theorem 65, TqS is the kernel of this matrix. If p ∈ H isimaginary then p has the form (ei, f + gi), so

qp = (−be− cf − dg + (ae+ cg − df)i, af − bg + de+ (ag + bf − ce)i)

and

a(−be− cf − dg) + b(ae+ cg − df) + c(af − bg + de) + d(ag + bf − ce) = 0.

42

Therefore qp ∈ TqS. This proves (1). For (2), let g = (a, b, c, d) and g′ = (e+fi, g+hi).We compute

DLg(g′) =

a −b −c −db a −d cc d a −bd −c b a

.Then

a −b −c −db a −d cc d a −bd −c b a

−feh−g

=

−(be+ af − dg + ch)ae− bf − cg − dhde− cf + bg + ah−(ce+ df + ag − bh)

,which shows that d(Lg)g′(X1|g′) = X1|gg′ , i.e. X1 is left-invariant. Similar calculationsshow that X2 and X3 are left-invariant.

Example 124. [Problem 8-9] Show by finding a counterexample that Proposition 8.19is false if we replace the assumption that F is a diffeomorphism by the weaker assump-tion that it is smooth and bijective.

Consider the smooth bijection ω : [0, 1) → S1 given by s 7→ e2πis and consider thesmooth vector field X given by x 7→ (1 − 2x) d/dt|x. There is no way of defining anF -related smooth vector field Y on S1, since the sign of Y1 is ambiguous.

Example 125. [Problem 8-10] Let M be the open submanifold of R2 where both xand y are positive, and let F : M → M be the map F (x, y) = (xy, y/x). Show that Fis a diffeomorphism, and compute F∗X and F∗Y , where

X = x∂

∂x+ y

∂

∂y, Y = y

∂

∂x.

The inverse

F−1(u, v) =

(√u

v,√uv

)is smooth, so F is a diffeomorphism. We compute

DF (x, y) =

[y x

−y/x2 1/x

]so that

DF (F−1(u, v)) =

[ √uv

√u/v

−v√v/u

√v/u

].

Therefore the coordinates of F∗X are given by[ √uv

√u/v

−v√v/u

√v/u

] [√u/v√uv

]=

[2u0

],

43

and the coordinates of F∗Y are given by[ √uv

√u/v

−v√v/u

√v/u

] [√uv0

]=

[uv−v2

].

So

F∗X = 2u∂

∂u, F∗Y = uv

∂

∂u− v2 ∂

∂v.

Example 126. [Problem 8-11] For each of the following vector fields on the plane,compute its coordinate representation in polar coordinates on the right half-plane(x, y) : x > 0.

(1) X = x∂

∂x+ y

∂

∂y.

(2) Y = x∂

∂x− y ∂

∂y.

(3) Z = (x2 + y2)∂

∂x.

Let M be the right half-plane. We have a diffeomorphism F : M → (0,∞) × (−π, π)

given by F (x, y) = (√x2 + y2, arctan(y/x)), with inverse F−1(r, θ) = (r cos θ, r sin θ).

Then

DF (x, y) =

[x/√x2 + y2 y/

√x2 + y2

−y/(x2 + y2) x/(x2 + y2)

],

so

DF (F−1(r, θ)) =

[cos θ sin θ

−r−1 sin θ r−1 cos θ

].

We have[cos θ sin θ

−r−1 sin θ r−1 cos θ

] [r cos θ r cos θ r2

r sin θ −r sin θ 0

]=

[r r cos 2θ r2 cos θ0 − sin 2θ −r sin θ

].

Therefore

X = r∂

∂r,

Y = r cos 2θ∂

∂r− sin 2θ

∂

∂θ,

Z = r2 cos θ∂

∂r− r sin θ

∂

∂θ.

Example 127. [Problem 8-12] Let F : R2 → RP2 be the smooth map F (x, y) = [x, y, 1],and let X ∈ X(R2) be defined by X = x∂/∂y − y∂/∂x. Prove that there is a vectorfield Y ∈ X(RP2) that is F -related to X, and compute its coordinate representation in

44

terms of each of the charts defined in Example 1.5. Let π : R3 \ 0 → RP2 be the

quotient map and define X : R3 \ 0 → TRP2 by

X(x,y,z) = dπ

(x∂

∂y

∣∣∣∣(x,y,z)

− y ∂

∂x

∣∣∣∣(x,y,z)

).

Since X is constant on the fibers of π, it descends to a vector field Y : RP2 → TRP2.

Let Ui ⊆ R3 \ 0, Ui ⊆ RP2 and ϕi : Ui → Rn be as in Example 1.5:

ϕi[x1, x2, x3] = (x1/xi, x2/xi, x3/xi),

ϕ−1i (u1, u2, u3) = [u1, . . . , ui−1, 1, ui+1, . . . , u3].

The coordinate representations of dπ and Y with respect to ϕ3 are given by

dπ(u1, u2, u3, v1, v2, v3) =

(u1

u3,u2

u3, 1,

v1

u3− u1v3

(u3)2,v2

u3− u2v3

(u3)2, 0

)and

Y (u1, u2, u3) = dπ

(u1 ∂

∂y

∣∣∣∣(u1,u2,1)

− u2 ∂

∂x

∣∣∣∣(u1,u2,1)

)= (u1, u2, 1,−u2, u1, 0).

On this chart, the coordinate representation of dF is

dF (u1, u2, v1, v2) = (u1, u2, 1, v1, v2, 0),

so for all (x, y) ∈ R2,

dF(x,y)(X(x,y)) = dF

(x∂

∂y

∣∣∣∣(x,y,z)

− y ∂

∂x

∣∣∣∣(x,y,z)

)= (x, y, 1,−y, x, 0)

= Y (x, y, 1).

The computations for ϕ1 and ϕ2 are similar. This shows that Y is F -related to X.

Theorem 128. [Problem 8-13] There is a smooth vector field on S2 that vanishes atexactly one point.

Proof. Let N = (0, 0, 1) be the north pole and let σ : S2\N → R2 be the stereographicprojection given by

σ(x1, x2, x3) = (x1, x2)/(1− x3),

σ−1(u1, u2) = (2u1, 2u2, |u|2 − 1)/(|u|2 + 1).

45

Let X be the 1st coordinate vector field on R2 and define a vector field Y on S2 by

Y (p) =

0, p = N,

((σ−1)∗X)p p 6= N.

This vector field vanishes at exactly one point. Let S = (0, 0,−1) be the south poleand let τ : S2 \ S → R2 be the stereographic projection given by

τ(x1, x2, x3) = (x1, x2)/(1 + x3),

τ−1(u1, u2) = (2u1, 2u2, 1− |u|2)/(|u|2 + 1).

Then(σ τ−1)(u1, u2) = (2u1, 2u2)/(2 |u|2),

and the coordinate representation of Y for (u1, u2) 6= (0, 0) with respect to τ is givenby

Y (u1, u2) = d(σ−1)(στ−1)(u1,u2)X(στ−1)(u1,u2)

= d(σ−1)(στ−1)(u1,u2)

(d

dx

∣∣∣∣(στ−1)(u1,u2)

)

=

(τ−1(u1, u2),

2(1− (u1)2 + (u2)2)

(1 + |u|2)2,− 4u1u2

(1 + |u|2)2,− 4u1

(1 + |u|2)2

),

which shows that Y is smooth on S2.

Theorem 129. [Problem 8-14] Let M be a smooth manifold with or without boundary,let N be a smooth manifold, and let f : M → N be a smooth map. Define F : M →M × N by F (x) = (x, f(x)). For every X ∈ X(M), there is a smooth vector field onM ×N that is F -related to X.

Proof. Let m = dimM and n = dimN . By Lemma 8.6, it suffices to show that there is asmooth vector field along F (M) that is F -related to X. By Proposition 5.4, F (M) is anembedded m-submanifold of M×N . Let p ∈ F (M), let (U,ϕ) be a slice chart for F (M)

containing p, let p = ϕ(p), and let U = ϕ(U) ⊆ Rm+n. By shrinking U , we may assume

that U is an open cube containing p, and that ϕ(F (M)∩U) = (a, b) ∈ Rm+n : b = 0.We identify TpU with U × Rm+n. Define a vector field Y on U by assigning Y(a,b) thecoefficients of dϕ(Xπ(ϕ−1(a,0))), where π : M×N →M is the canonical projection. ThenY is smooth, and (ϕ−1)∗Y is a smooth vector field defined on U that agrees with X onF (M) ∩ U .

Theorem 130. [Problem 8-15] Suppose M is a smooth manifold and S ⊆ M is anembedded submanifold with or without boundary. Given X ∈ X(S), there is a smoothvector field Y on a neighborhood of S in M such that X = Y |S. Every such vector fieldextends to all of M if S is properly embedded.

46

Proof. As in Theorem 82.

Example 131. [Problem 8-16] For each of the following pairs of vector fields X, Ydefined on R3, compute the Lie bracket [X, Y ].

(1) X = y∂

∂z− 2xy2 ∂

∂y; Y =

∂

∂y.

(2) X = x∂

∂y− y ∂

∂x; Y = y

∂

∂z− z ∂

∂y.

(3) X = x∂

∂y− y ∂

∂x; Y = x

∂

∂y+ y

∂

∂x.

For (1), we have∂X2

∂x= −2y2,

∂X2

∂y= −4xy,

∂X3

∂y= 1,

so

[X, Y ] = 4xy∂

∂y− ∂

∂z.

For (2), we have

∂X1

∂y= −1,

∂X2

∂x= 1,

∂Y 2

∂z= −1,

∂Y 3

∂y= −1,

so

[X, Y ] = z∂

∂x− x ∂

∂z.

For (3), we have

∂X1

∂y= −1,

∂X2

∂x= 1,

∂Y 1

∂y= 1,

∂Y 2

∂x= 1,

so

[X, Y ] = 2x∂

∂x− 2y

∂

∂y.

Theorem 132. [Problem 8-17] Let M and N be smooth manifolds. Given vector fieldsX ∈ X(M) and Y ∈ X(N), we can define a vector field X ⊕ Y on M ×N by

(X ⊕ Y )(p,q) = (Xp, Yq),

where we think of the right-hand side as an element of TpM ⊕ TqN , which is naturallyidentified with Tp,q(M ×N) as in Proposition 3.14. Then X ⊕ Y is smooth if X and Yare smooth, and [X1 ⊕ Y1, X2 ⊕ Y2] = [X1, X2]⊕ [Y1, Y2].

Proof. The smoothness of X⊕Y follows easily from Proposition 8.1. If f ∈ C∞(M×N)then

[X1 ⊕ Y1, X2 ⊕ Y2](p,q)f = (X1 ⊕ Y1)(p,q)(X2 ⊕ Y2)f − (X2 ⊕ Y2)(p,q)(X1 ⊕ Y1)f

47

= (X1|pX2fq, Y1|qY2fp)− (X2|pX1fq, Y2|qY1fp)

= ([X1, X2]fq, [Y1, Y2]fp)

= ([X1, X2]⊕ [Y1, Y2])f,

where fq ∈ C∞(M) is the map p 7→ f(p, q) and fp ∈ C∞(N) is the map q 7→ f(p, q).

Theorem 133. [Problem 8-18] Suppose F : M → N is a smooth submersion, where Mand N are positive-dimensional smooth manifolds. Given X ∈ X(M) and Y ∈ X(N),we say that X is a lift of Y if X and Y are F -related. A vector field V ∈ X(M) issaid to be vertical if V is everywhere tangent to the fibers of F (or, equivalently, if Vis F -related to the zero vector field on N).

(1) If dimM = dimN , then every smooth vector field on N has a unique lift.(2) If dimM 6= dimN , then every smooth vector field on N has a lift, but it is not

unique.(3) Assume in addition that F is surjective. Given X ∈ X(M), X is a lift of a

smooth vector field on N if and only if dFp(Xp) = dFq(Xq) whenever F (p) =F (q). If this is the case, then X is a lift of a unique smooth vector field.

Proof. Let m = dimM and n = dimN . If m = n then F is a local diffeomorphism,and if Y ∈ X(N) then we can define a smooth vector field X ∈ X(M) by

Xp = (dFp)−1(YF (p)).

This is clearly the only vector field that satisfies dFp(Xp) = YF (p), and smoothnessfollows from Theorem 41 and the fact that dF is a local diffeomorphism. This proves(1). Now suppose Y ∈ X(N) and m 6= n, i.e. m > n. Let p ∈ M , let q = F (p), andchoose smooth coordinates (xi) centered at p and (yi) centered at q in which F hasthe coordinate representation π(x1, . . . , xm) = (x1, . . . , xn). If ε is a sufficiently smallnumber, the coordinate cube

Cp =x :∣∣xi∣∣ < ε for i = 1, . . . ,m

is a neighborhood of p whose image under F is the cube

C ′p =y :∣∣yi∣∣ < ε for i = 1, . . . , n

.

Define a smooth vector field Xp on Cp by setting X ip(x) = Y i(F (x)) for i = 1, . . . , n

and X ip(x) = 0 for i = n + 1, . . .m. Then dFx(Xp|x) = YF (x) for all x ∈ Cp. Choose a

partition of unity ψp subordinate to the open cover Cp. Then

X =∑p∈M

ψpXp

48

is a smooth vector field on M (taking ψpXp = 0 on M \ suppψp), and

dFx(Xx) = dFx

(∑p∈M

ψp(x)Xp|x

)=∑p∈M

ψp(x)dFx(Xp|x)

=∑p∈M

ψp(x)YF (x)

= YF (x)

for all x ∈ M . This proves (2). For (3), F is a quotient map. If X is a lift of someY ∈ X(N) then for all p, q ∈M ,

dFp(Xp) = YF (p) = YF (q) = dFq(Xq).

Conversely, if dFp(Xp) = dFq(Xq) whenever F (p) = F (q) then dF X is constant on thefibers of F , so it descends to a smooth vector field Y on N satisfying dF X = Y F ,i.e. dFp(Xp) = YF (p) for all p ∈M .

Theorem 134. [Problem 8-20] Let A ⊆ X(R3) be the subspace spanned by X, Y, Z,where

X = y∂

∂z− z ∂

∂y, Y = z

∂

∂x− x ∂

∂z, Z = x

∂

∂y− y ∂

∂x.

Then A is a Lie subalgebra of X(R3) which is isomorphic to R3 with the cross product.

Proof. Since X, Y, Z is linearly independent, it is a basis for A. Simple computationssimilar to those in Example 131 show that A is closed under brackets. Define anisomorphism ϕ : A → R3 by setting ϕX = (1, 0, 0), ϕY = (0, 1, 0), ϕZ = (0, 0, 1), andextending linearly. It is easy to check that ϕ is a Lie algebra isomorphism. For example,

ϕ[Z,X] = ϕY = [(0, 0, 1), (1, 0, 0)] = [ϕZ, ϕX].

Theorem 135. [Problem 8-21] Up to isomorphism, there are exactly one 1-dimensionalLie algebra and two 2-dimensional Lie algebras. All three algebras are isomorphic toLie subalgebras of gl(2,R).

Proof. Any 1-dimensional Lie algebra is abelian, and is isomorphic to the subalgebrarI2 : r ∈ R of gl(2,R). Let A be a 2-dimensional Lie algebra and let x, y be a basisfor A. If A is abelian then it is isomorphic to the subalgebra[

r1 00 r2

], r1, r2 ∈ R

49

of gl(2,R). Otherwise, we have [x, y] = ax + by for some a, b ∈ R, not both zero.Assume without loss of generality that b 6= 0. (If b = 0, then we can swap x and y.)Define an homomorphism ϕ : A→ gl(2,R) by setting

ϕx =

[1 00 1 + b

], ϕy =

[−ab−1 0

1 −ab−1(1 + b)

]and extending linearly. Then ϕ is an isomorphism onto its image, and it is easy toverify that ϕ[x, y] = [ϕx, ϕy].

Theorem 136. [Problem 8-22] Let A be any algebra over R. A derivation of A is alinear map D : A → A satisfying D(xy) = (Dx)y + x(Dy) for all x, y ∈ A. If D1 andD2 are derivations of A, then [D1, D2] = D1 D2 −D2 D1 is also a derivation. Theset of derivations of A is a Lie algebra with this bracket operation.

Proof. Identical to Lemma 8.25.

Theorem 137. [Problem 8-23]

(1) Given Lie algebras g and h, the direct sum g⊕h is a Lie algebra with the bracketdefined by

[(X, Y ), (X, Y ′)] = ([X,X ′], [Y, Y ′]).

(2) Suppose G and H are Lie groups. Then Lie(G×H) is isomorphic to Lie(G)⊕Lie(H).

Proof. We have isomorphisms ε : Lie(G ×H) → Te(G ×H), ε1 : Lie(G) → Te(G) andε2 : Lie(H)→ Te(H). Thus we have an isomorphism

ε′ : Lie(G)⊕ Lie(H)→ Te(G)⊕ Te(H) ∼= Te(G×H),

and it remains to show that ε−1 ε′ respects brackets. Let (X1, X2), (Y1, Y2) ∈ Lie(G)⊕Lie(H); then

(ε−1 ε′)[(X1, X2), (Y1, Y2)](g,h) = ((ε−1 ε′)([X1, Y1], [X2, Y2]))(g,h)

= (ε−1([X1, Y1]e, [X2, Y2]e))(g,h)

= d(L(g,h))(e,e)([X1, Y1]⊕ [X2, Y2])(e,e)

= d(L(g,h))(e,e)[X1 ⊕X2, Y1 ⊕ Y2](g,h)

= [X1 ⊕X2, Y1 ⊕ Y2](g,h)

= [(ε−1 ε′)(X1, X2), (ε−1 ε′)(Y1, Y2)](g,h)

since [X1 ⊕X2, Y1 ⊕ Y2] is left-invariant (see Theorem 132).

Theorem 138. [Problem 8-24] Suppose G is a Lie group and g is its Lie algebra. Avector field X ∈ X(G) is said to be right-invariant if it is invariant under all righttranslations.

50

(1) The set g of right-invariant vector fields on G is a Lie subalgebra of X(G).(2) Let i : G → G denote the inversion map i(g) = g−1. The pushforward i∗ :

X(G)→ X(G) restricts to a Lie algebra isomorphism from g to g.

Proof. Part (1) follows from Corollary 8.31. If X ∈ g then for all g, g′ ∈ G we have

d(Rg)g′((i∗X)g′) = (d(Rg)g′ dii−1(g′))(Xi−1(g′))

= d(Rg i)(g′)−1(X(g′)−1)

= d(i Lg−1)(g′)−1(X(g′)−1)

= di(g′g)−1(d(Lg−1)(g′)−1X(g′)−1)

= di(g′g)−1(X(g′g)−1)

= (i∗X)g′g.

Therefore i∗(g) ⊆ g. A similar calculation shows that i∗(g) ⊆ g, so (i∗)−1 = i∗ and i∗ is

an isomorphism. Corollary 8.31 shows that i∗ is a Lie algebra isomorphism.

Theorem 139. [Problem 8-25] If G is an abelian Lie group, then Lie(G) is abelian.

Proof. If G is abelian then the inversion map i : G → G is a Lie group isomorphism.The induced Lie algebra isomorphism is given by

(i∗X)g = d(Lg)e(die(Xe))

= d(Lg)e(−Xe)

= −d(Lg)eXe

= −Xg,

so i∗X = −X for all X ∈ Lie(G). Then

[X, Y ] = [−X,−Y ] = [i∗X, i∗Y ] = i∗[X, Y ] = −[X, Y ],

so [X, Y ] = 0 for all X, Y ∈ Lie(G). This proves that Lie(G) is abelian.

Theorem 140. [Problem 8-26] Suppose F : G → H is a Lie group homomorphism.The kernel of F∗ : Lie(G)→ Lie(H) is the Lie algebra of kerF (under the identificationof the Lie algebra of a subgroup with a Lie subalgebra as in Theorem 8.46).

Proof. Consider ι∗X for some X ∈ kerF , where ι : kerF → G is the inclusion map.Then F∗ι∗X = (F ι)∗X = 0 since F ι = 0. This shows that ι∗(Lie(kerF )) ⊆ kerF∗.Conversely, suppose F∗Y = 0 for some Y ∈ Lie(G). By definition we have dFe(Ye) = 0,so Ye ∈ ker(dFe) = Te(kerF ) by Theorem 65. By Theorem 8.46, this shows thatkerF∗ ⊆ ι∗(Lie(kerF )).

51

Theorem 141. [Problem 8-27] Let G and H be Lie groups, and suppose F : G →H is a Lie group homomorphism that is also a local diffeomorphism. The inducedhomomorphism F∗ : Lie(G)→ Lie(H) is an isomorphism of Lie algebras.

Proof. Since F is a local diffeomorphism, dFe is bijective. If F∗X = 0 then dFe(Xe) = 0,so Xe = 0 and X = 0. Suppose Y ∈ Lie(H). Let ε : Lie(G) → TeG be the canonicalisomorphism. Then

(F∗ε−1(dFe)

−1Ye)e = dFe((ε−1(dFe)

−1Ye)e)

= dFe((dFe)−1Ye)

= Ye,

so F∗(ε−1(dFe)

−1Ye) = Y . This proves that F∗ is a bijection.

Theorem 142. [Problem 8-28] Considering det : GL(n,R) → R∗ as a Lie group ho-momorphism, its induced Lie algebra homomorphism is tr : gl(n,R)→ R.

Proof. This follows immediately from Theorem 98.

Example 143. [Problem 8-29] Theorem 8.46 implies that the Lie algebra of any Liesubgroup of GL(n,R) is canonically isomorphic to a subalgebra of gl(n,R), with asimilar statement for Lie subgroups of GL(n,C). Under this isomorphism, show that

Lie(SL(n,R)) ∼= sl(n,R),

Lie(SO(n)) ∼= o(n),

Lie(SL(n,C)) ∼= sl(n,C),

Lie(U(n)) ∼= u(n),

Lie(SU(n)) ∼= su(n),

where

sl(n,R) = A ∈ gl(n,R) : trA = 0,o(n) = A ∈ gl(n,R) : AT + A = 0,

sl(n,C) = A ∈ gl(n,C) : trA = 0,u(n) = A ∈ gl(n,C) : A∗ + A = 0,su(n) = u(n) ∩ sl(n,C).

We have SL(n,R) = ker det, so Lie(SL(n,R)) ∼= ker det∗ = ker tr = sl(n,R). The samecomputation holds for SL(n,C). The others follow by considering the kernel of the mapA 7→ ATA (or A 7→ A∗A in the complex case).

52

Example 144. [Problem 8-30] Show by giving an explicit isomorphism that su(2) ando(3) are isomorphic Lie algebras, and that both are isomorphic to R3 with the crossproduct.

We have an isomorphism from R3 to o(3) given byabc

7→ 0 c −b−c 0 ab −a 0

.We also have an isomorphism from R3 to su(2) given byab

c

7→ 1√2

[ai b+ ci

−b+ ci −ai

].

Theorem 145. [Problem 8-31] Let g be a Lie algebra. A linear subspace h ⊆ g is calledan ideal in g if [X, Y ] ∈ h whenever X ∈ h and Y ∈ g.

(1) If h is an ideal in g, then the quotient space g/h has a unique Lie algebrastructure such that the projection π : g→ g/h is a Lie algebra homomorphism.

(2) A subspace h ⊆ g is an ideal if and only if it is the kernel of a Lie algebrahomomorphism.

Proof. Define a bracket on g/h by [X + h, Y + h] = [X, Y ] + h. If X ′ + h = X + h andY ′ + h = Y + h then

[X ′, Y ′] = [X ′ +X −X ′, Y ′] = [X, Y ′ + Y − Y ′] = [X, Y ]

since X −X ′ ∈ h and Y − Y ′ ∈ h, which shows that the bracket is well-defined. It isclearly the unique bracket making π a Lie algebra homomorphism. If h ⊆ g is an idealthen it is the kernel of the projection π : g → g/h. Conversely, if f is a Lie algebrahomomorphism then ker f is an ideal since X ∈ ker f implies that

f [X, Y ] = [fX, fY ] = 0.

Chapter 9. Integral Curves and Flows

Theorem 146. [Exercise 9.37] Suppose v ∈ Rn and W is a smooth vector field on anopen subset of Rn. The directional derivative

DvW (p) =d

dt

∣∣∣∣t=0

Wp+tv

is equal to (LVW )p, where V is the vector field V = vi∂/∂xi with constant coefficientsin standard coordinates.

53

Proof. The flow of V is given by θ(t, x) = x+ tv. Therefore

(LVW )p =d

dt

∣∣∣∣t=0

d(θ−t)θt(p)(Wθt(p))

=d

dt

∣∣∣∣t=0

d(θ−t)p+tv(Wp+tv)

=d

dt

∣∣∣∣t=0

Wp+tv

since θ−t(x) = x− tv is a translation of x.

Theorem 147. [Problem 9-1] Suppose M is a smooth manifold, X ∈ X(M), and γ isa maximal integral curve of X.

(1) We say that γ is periodic if there is a number T > 0 such that γ(t+ T ) = γ(t)for all t ∈ R. Exactly one of the following holds:(a) γ is constant.(b) γ is injective.(c) γ is periodic and nonconstant.

(2) If γ is periodic and nonconstant, then there exists a unique positive number T(called the period of γ) such that γ(t) = γ(t′) if and only if t − t′ = kT forsome k ∈ Z.

(3) The image of γ is an immersed submanifold of M , diffeomorphic to R, S1, orR0.

Proof. If γ is constant then (b) and (c) cannot hold, so we assume that γ is nonconstant.If γ is not injective then γ(t0) = γ(t1) for some points t0 < t1, and γ is defined on atleast (t0− ε, t1 + ε) for some ε > 0. Let T = t1− t0. Then γ and t 7→ γ(t1 + t) are bothintegral curves of X starting at γ(t0), so Theorem 9.12 shows that γ must be definedon at least (t0− ε, t1 +T + ε). By induction, t0 +kT is in the domain of γ for all k ∈ Z,so γ is defined on R and has period T . This proves (1). For (2), let A be the set of allT > 0 such that γ(t + T ) = γ(t) for all t ∈ R, which is nonempty since γ is periodic.If we can show that A is closed, then T0 = inf A ∈ A is the period of γ. But if T /∈ Athen γ(t+ T ) 6= γ(t) for some t ∈ R, so

γ−1(M \ γ(t))− t = s− t : γ(s) 6= γ(t)is a neighborhood of T contained in R\A. For (3), if γ is constant then the image of γ isdiffeomorphic to R0. Otherwise, Proposition 9.21 shows that γ is a smooth immersion.If γ is injective then Proposition 5.18 shows that the image of γ is diffeomorphic toR, and if γ is periodic then it descends to a smooth injective immersion from S1 to M(using a suitable smooth covering of S1 by R), which is a diffeomorphism onto its imageby Proposition 5.18.

54

Theorem 148. [Problem 9-2] Suppose M is a smooth manifold, S ⊆M is an immersedsubmanifold, and V is a smooth vector field on M that is tangent to S.

(1) For any integral curve γ of V such that γ(t0) ∈ S, there exists ε > 0 such thatγ ((t0 − ε, t0 + ε)) ⊆ S.

(2) If S is properly embedded, then every integral curve that intersects S is containedin S.

(3) (2) need not hold if S is not closed.

Proof. Let x = γ(t0) ∈ S. By Proposition 8.23, there is a smooth vector field V |S on Sthat is ι-related to V . Let α : I → S be a maximal integral curve of V |S starting at x,where I contains (−ε, ε) for some ε > 0. Then

(ι α)′(0) = dιx(α′(0)) = dιx((V |S)x) = Vx,

which shows that ι α is an integral curve of V starting at x. By Theorem 9.12, γ isdefined on J = (t0 − ε, t0 + ε) and γ(t) = (ι α)(t− t0) ∈ S for t ∈ J . This proves (1).Now suppose that S is properly embedded and let γ be an integral curve that intersectsS, i.e. γ(t0) ∈ S for some t0. Let α be defined as above. By Theorem 9.12, γ is definedon at least I + t0 = (a, b) and γ(t) ∈ S for all t ∈ (a, b). Suppose γ can be extended toan open interval J larger than (a, b). Since S is closed, γ(b) ∈ S by continuity. Then(1) shows that γ ((b− ε, b+ ε)) ⊆ S for some ε > 0, which contradicts the maximalityof α. This proves (2). For (3), let S be the open ball of radius 1 in R and let V be the1st coordinate vector field on R. Then θ : R × R → R given by θ(t, x) = x + t is anintegral curve of V that intersects S, but θ is not contained in S.

Example 149. [Problem 9-3] Compute the flow of each of the following vector fieldson R2:

(1) V = y∂

∂x+

∂

∂y.

(2) W = x∂

∂x+ 2y

∂

∂y.

(3) X = x∂

∂x− y ∂

∂y.

(4) Y = y∂

∂x+ x

∂

∂y.

Let γ(t) = (x(t), y(t)) be a curve. We have the differential equations

x′(t) = y(t), y′(t) = 1,

x′(t) = x(t), y′(t) = 2y(t),

x′(t) = x(t), y′(t) = −y(t)

x′(t) = y(t), y′(t) = x(t).

55

Therefore the flows (in this order) are

θt(x, y) =

(x+ ty +

1

2t2, y + t

),

θt(x, y) = (xet, ye2t),

θt(x, y) = (xet, ye−t),

θt(x, y) =1

2

(et(x+ y) + e−t(x− y), et(x+ y)− e−t(x− y)

).

Theorem 150. [Problem 9-4] For any integer n ≥ 1, define a flow on the odd-dimensional sphere S2n−1 ⊆ Cn by θ(t, z) = eitz. The infinitesimal generator of θis a smooth nonvanishing vector field on S2n−1.

Proof. This is obvious.

Theorem 151. [Problem 9-5] Suppose M is a smooth, compact manifold that admitsa nowhere vanishing smooth vector field. There exists a smooth map F : M → M thatis homotopic to the identity and has no fixed points.

Proof. Let V be a nowhere vanishing smooth vector field on M . Since M is compact,Corollary 9.17 shows that there is a global flow θ : R ×M → M of V . By Theorem9.22, each p has a neighborhood Up in which V has the coordinate representation∂/∂s1. Choosing a sufficiently small neighborhood Vp of p, there is some Tp > 0 suchthat θt(x) = x + (t, 0, . . . , 0) in local coordinates for all 0 ≤ t ≤ Tp and x ∈ Vp.Since Vp : p ∈M is an open cover of M , there is a finite subcover Vp1 , . . . , Vpn. LetT = min Tp1 , . . . , Tpn. Then θT has no fixed points, and the map H : M × I → Mgiven by (x, t) 7→ θ(tT, x) is a homotopy from the identity to θT .

Theorem 152. [Problem 9-6] Suppose M is a smooth manifold and V ∈ X(M). Ifγ : J → M is a maximal integral curve of V whose domain J has a finite least upperbound b, then for any t0 ∈ J , γ ([t0, b)) is not contained in any compact subset of M .

Proof. Let θ be the flow of V . Let bn be an increasing sequence contained in [t0, b)and converging to b. Since γ ([t0, b)) is contained in a compact set E ⊆ M , there is asubsequence γ(bnk) of γ(bn) that converges to a point p in E. Choose some ε > 0and a neighborhood U of p such that θ is defined on (−2ε, 2ε) × U , and choose aninteger m such that bm ∈ (b− ε, b) and γ(bm) ∈ U . Define γ1 : [t0, b+ ε)→M by

γ1(t) =

γ(t), t0 ≤ t < b,

θ(γ(bm))(t− bm) b ≤ t < b+ ε.

For all t ∈ (bm, b) we have

θ(γ(bm))(t− bm) = θt−bm(γ(bm)) = γ(t),

56

so γ1 is smooth because γ and t 7→ θ(γ(bm))(t − bm) agree where they overlap. But γ1

extends γ at b, which contradicts the maximality of γ.

Theorem 153. [Problem 9-7] Let M be a connected smooth manifold. The group ofdiffeomorphisms of M acts transitively on M : that is, for any p, q ∈ M , there is adiffeomorphism F : M →M such that F (p) = q.

Proof. First assume that M = Bn. By applying Lemma 8.6 to the map x 7→ q − pdefined on the line segment between p and q, we have a compactly supported smoothvector field on Bn whose flow θ satisfies θ1(p) = q. Theorem 9.16 shows that θ1 is adiffeomorphism on Bn, which proves the result for M = Bn. The general case followsfrom Theorem 29 and the fact that every point of M is contained in a coordinateball.

Theorem 154. [Problem 9-8] Let M be a smooth manifold and let S ⊆M be a compactembedded submanifold. Suppose V ∈ X(M) is a smooth vector field that is nowheretangent to S. There exists ε > 0 such that the flow of V restricts to a smooth embeddingΦ : (−ε, ε)× S →M .

Proof. Let θ : D → M be the flow of V and let Φ = θO where O = (R × S) ∩ D. Letδ : S → R be as in Theorem 9.20. Since S is compact, δ attains a minimum α on S, andΦ : (−α, α)×S →M is a smooth immersion. Let ε = α/2. Then [−ε, ε]×S is compact,so Φ : [−ε, ε]× S →M is a smooth embedding. Therefore Φ : (−ε, ε)× S →M is alsoa smooth embedding.

Theorem 155. [Problem 9-9] Suppose M is a smooth manifold and S ⊆ M is anembedded hypersurface (not necessarily compact). Suppose further that there is a smoothvector field V defined on a neighborhood of S and nowhere tangent to S. Then S has aneighborhood in M diffeomorphic to (−1, 1)× S, under a diffeomorphism that restrictsto the obvious identification 0 × S ≈ S.

Proof. Using Theorem 9.20, we have that Φ|Oδ is a diffeomorphism onto an open sub-manifold of M . So it remains to show that Oδ = (t, p) ∈ O : |t| < δ(p) is diffeomor-phic to (−1, 1)× S. Let ϕ : (−1, 1)× S → Oδ be given by (t, p) 7→ (tδ(p), p); then ϕ issmooth, and its inverse ϕ−1(t, p) = (t/δ(p), p) is also smooth.

Example 156. [Problem 9-10] For each vector field in Example 149, find smoothcoordinates in a neighborhood of (1, 0) for which the given vector field is a coordinatevector field.

We only consider the first vector field. We have V(1,0) = ∂/∂y, so we can take S to bethe x-axis, parametrized by X(s) = (s, 0). Therefore

Ψ(t, s) = θt(s, 0) =

(s+

1

2t2, t

),

57

and

Ψ−1(x, y) =

(x− 1

2y2, y

);

DΨ−1(x, y) =

[1 −y0 1

].

In these coordinates, we have

V = y∂

∂x+

∂

∂y

= s∂

∂s+

(−s ∂

∂s+∂

∂t

)=

∂

∂t.

Chapter 10. Vector Bundles

Theorem 157. [Exercise 10.1] Suppose E is a smooth vector bundle over M . Theprojection map π : E →M is a surjective smooth submersion.

Proof. Let v ∈ E and let Φ : π−1(U) → U × Rk be a smooth local trivialization of Esuch that π(v) ∈ U . Then πU Φ = π, so dπv = d(πU)π(v) dΦv is surjective since dΦv

and d(πU)π(v) are both surjective.

Theorem 158. [Exercise 10.9] The zero section ζ of every vector bundle π : E → Mis continuous, and the zero section of every smooth vector bundle is smooth.

Proof. Let p ∈M and let Φ : π−1(U)→ U ×Rk be a local trivialization of E such thatp ∈ U . We have (Φ ζ)(x) = (x, 0) for all x ∈ U , so Φ ζ|U is continuous and ζ|U iscontinuous since Φ is a homeomorphism. But p was arbitrary, so ζ is continuous. Asimilar argument holds if π is a smooth vector bundle.

Theorem 159. [Exercise 10.11] Let π : E →M be a smooth vector bundle.

(1) If σ, τ ∈ Γ(E) and f, g ∈ C∞(M), then fσ + gτ ∈ Γ(E).(2) Γ(E) is a module over the ring C∞(M).

Proof. Let Φ : π−1(U)→ U × Rk be a local trivialization of E; then

(Φ (fσ + gτ))(x) = (x, f(x)πRk(σ(x)) + g(x)πRk(τ(x))) ,

so fσ + gτ is smooth on U . Therefore fσ + gτ is smooth.

58

Theorem 160. [Exercise 10.13] Let π : E → M be a smooth vector bundle over asmooth manifold M with or without boundary. Suppose A is a closed subset of M ,and σ : A → E is a section of E|A that is smooth in the sense that σ extends to asmooth local section of E in a neighborhood of each point. For each open subset U ⊆Mcontaining A, there exists a global smooth section σ ∈ Γ(E) such that σ|A = σ andsupp σ ⊆ U .

Proof. For each p ∈ A, choose a neighborhood Wp of p and a local section σp : Wp → Eof E that agrees with σ on Wp ∩ A. Replacing Wp by Wp ∩ U we may assume thatWp ⊆ U . The family of sets Wp : p ∈ A ∪ M \ A is an open cover of M . Letψp : p ∈ A ∪ ψ0 be a smooth partition of unity subordinate to this cover, withsuppψp ⊆ Wp and suppψ0 ⊆M \ A.

For each p ∈ A, the product ψpσp is smooth on Wp by Theorem 159, and has a smoothextension to all of M if we interpret it to be zero on M \ suppψp. (The extendedfunction is smooth because the two definitions agree on the open subset Wp \ suppψpwhere they overlap.) Thus we can define σ : M → E by

σ(x) =∑p∈A

ψp(x)σp(x).

Because the collection of supports suppψp is locally finite, this sum actually has onlya finite number of nonzero terms in a neighborhood of any point of M , and thereforedefines a smooth function. If x ∈ A, then ψ0(x) = 0 and σp(x) = σ(x) for each p suchthat ψp(x) 6= 0, so

σ(x) =∑p∈A

ψp(x)σ(x) =

(ψ0(x) +

∑p∈A

ψp(x)

)σ(x) = σ(x),

so σ is indeed an extension of σ. It follows from Lemma 1.13(b) that

supp σ =⋃p∈A

suppψp =⋃p∈A

suppψp ⊆ U.

Theorem 161. [Exercise 10.14] Let π : E → M be a smooth vector bundle. Everyelement of E is in the image of a smooth global section.

Proof. Let v ∈ E and let p = π(v). The assignment p 7→ v is a section of E|p, whichis easily seen to be smooth in the sense of Theorem 160.

Theorem 162. [Exercise 10.16] Suppose π : E →M is a smooth vector bundle of rankk.

59

(1) If (σ1, . . . , σm) is a linearly independent m-tuple of smooth local sections of Eover an open subset U ⊆ M , with 1 ≤ m < k, then for each p ∈ U there existsmooth sections σm+1, . . . , σk defined on some neighborhood V of p such that(σ1, . . . , σk) is a smooth local frame for E over U ∩ V .

(2) If (v1, . . . , vm) is a linearly independent m-tuple of elements of Ep for somep ∈ M , with 1 ≤ m ≤ k, then there exists a smooth local frame (σi) for E oversome neighborhood of p such that σi(p) = vi for i = 1, . . . ,m.

(3) If A ⊆ M is a closed subset and (τ1, . . . , τk) is a linearly independent k-tupleof sections of E|A that are smooth in the sense described in Theorem 160, thenthere exists a smooth local frame (σ1, . . . , σk) for E over some neighborhood ofA such that σi|A = τi for i = 1, . . . , k.

Proof. Choose a smooth local trivialization Φ : ϕ−1(V )→ V ×Rk such that p ∈ V . Byshrinking V , we may assume there is a coordinate map ϕ : V → Rn and that V ⊆ U ;let (xi) be the coordinates. Let vi = (πRk Φ σi)(p) for i = 1, . . . ,m, and completev1, . . . , vm to a basis v1, . . . , vk of Rk. For each i = m + 1, . . . , k, let σi : V → Ebe the local section of E defined by σi(x) = Φ−1(x, vi). Let W be the subspace of Rk

spanned by vm+1, . . . , vk; then F = Φ−1(V × (Rn \W )) is open, so Vi = σ−1i (F ) is a

neighborhood of p for each i = 1, . . . ,m. Let V = V ∩ V1 ∩ · · · ∩ Vm. Then σ1, . . . , σkis a smooth local frame for E on V since (πRk Φ σi)(x) /∈ W for all x ∈ V andi = 1, . . . , k. This proves (1). Part (2) follows by choosing constant coefficient localsections σi such that σi(p) = vi on a sufficiently small neighborhood of p. For (3), applyTheorem 160 to each τi and restrict to a sufficiently small neighborhood of A.

Theorem 163. [Exercise 10.27] A smooth rank-k vector bundle over M is smoothlytrivial if and only if it is smoothly isomorphic over M to the product bundle M × Rk.

Proof. If π : E →M is smoothly trivial then there is a trivialization Φ : E →M ×Rk.This map satisfies πM Φ = π, so E and M ×Rk are smoothly isomorphic over M . Theconverse is similar.

Theorem 164. [Exercise 10.31] Given a smooth vector bundle π : E → M and asmooth subbundle D ⊆ E, the inclusion map ι : D → E is a smooth bundle homomor-phism over M .

Proof. This is simply the condition π ι = π|D.

Theorem 165. [Problem 10-1] The Mobius bundle E is not trivial.

Proof. E is not homeomorphic to S1 × R. Removing the center circle S1 × 0 fromS1 × R leaves a disconnected space, but removing the center circle from E leaves aconnected space.

60

Theorem 166. [Problem 10-2] Let E be a vector bundle over a topological space M .The projection map π : E →M is a homotopy equivalence.

Proof. Let ζ : M → E be the zero section. Then π ζ = IdM , and H : E × I → Edefined by H(v, t) = t IdE is a homotopy from ζ π to IdE. This shows that π is ahomotopy equivalence.

Theorem 167. [Problem 10-3] Let VB denote the category whose objects are smoothvector bundles and whose morphisms are smooth bundle homomorphisms, and let Diffdenote the category whose objects are smooth manifolds and whose morphisms aresmooth maps. The assignment M 7→ TM , F 7→ dF defines a covariant functor fromDiff to VB, called the tangent functor.

Proof. This follows immediately from Corollary 3.22.

Theorem 168. [Problem 10-4] The map τ : U ∩ V → GL(k,R) in Lemma 10.5 issmooth.

Proof. We know that σ : (U ∩ V ) × Rk → Rk given by σ(p, v) = τ(p)v is smooth.Let e1, . . . , ek be the standard basis for Rk. The (i, j) entry of τ(p) is πi(τ(p)ej) =πi(σ(p, ej)), where πi : Rk → R is the projection onto the ith coordinate. This map issmooth as a function of p, so τ is smooth.

Theorem 169. [Problem 10-5] Let π : E →M be a smooth vector bundle of rank k overa smooth manifold M with or without boundary. Suppose that Uαα∈A is an open coverof M , and for each α ∈ A we are given a smooth local trivialization Φα : π−1(Uα) →Uα ×Rk of E. For each α, β ∈ A such that Uα ∩ Uβ 6= ∅, let ταβ : Uα ∩ Uβ → GL(k,R)be the transition function defined by (10.3). The following identity is satisfied for allα, β, γ ∈ A:

ταβ(p)τβγ(p) = ταγ(p), p ∈ Uα ∩ Uβ ∩ Uγ.

Proof. This follows from the fact that

(p, ταγ(p)v) = (Φα Φ−1γ )(p, v)

= (Φα Φ−1β Φβ Φ−1

γ )(p, v)

= (p, ταβ(p)τβγ(p)v).

Theorem 170. [Problem 10-6] Let M be a smooth manifold with or without boundary,and let Uαα∈A be an open cover of M . Suppose for each α, β ∈ A we are given asmooth map ταβ : Uα ∩ Uβ → GL(k,R) such that

ταβ(p)τβγ(p) = ταγ(p), p ∈ Uα ∩ Uβ ∩ Uγ

61

is satisfied for all α, β, γ ∈ A. Then there is a smooth rank-k vector bundle E → Mwith smooth local trivializations Φα : π−1(Uα) → Uα × Rk whose transition functionsare the given maps ταβ.

Proof. Note that ταα = IdRk for all α ∈ A. Let F =∐

α∈A Uα×Rk and let iα : Uα×Rk →F be the canonical injection for each α ∈ A. Define an equivalence relation ∼ on F bydeclaring that (p, v)α ∼ (p′, v′)β if and only if p = p′ and ταβ(p)v′ = v. This relation issymmetric because

ταβ(p)τβα(p) = ταα(p) = IdRk ,

so ταβ(p)v′ = v implies that v′ = τβα(p)v. Similarly, the relation is transitive becauseταβ(p)v′ = v and τβγ(p)v

′′ = v′ implies that

ταγ(p)v′′ = ταβ(p)τβγ(p)v

′′ = ταβ(p)v′ = v.

Let E = F/ ∼ and let q : F → E be the quotient map. Let Ep = q(p × Rk

)for each

p ∈ M ; we want to give Ep a vector space structure. Define m : R × q−1(Ep) → Epby m(r, (p, v)α) = q((p, rv)α). If (p, v)α ∼ (p, v′)β then ταβ(p)v′ = v and ταβ(p)rv′ =rταβ(p)v′ = rv. Therefore m descends to a continuous map m : R×Ep → Ep, which isscalar multiplication in Ep. A similar construction shows that vector addition can bedefined on Ep.

Let π : F →M be given by (p, v) 7→ p; then π descends to a continuous map π : E →Msatisfying π q = π. For each α ∈ A, define Φα : π−1(Uα) → Uα × Rk as the uniquebijection satisfying Φα q iα = IdUα×Rk . We now check conditions (ii) and (iii) inLemma 10.6. For (ii), it is clear that Φα|Ep is linear, so it is an isomorphism. For (iii),we have

(Φα Φ−1β )(p, v) = Φα(q((p, v)β))

= Φα(q((p, ταβ(p)v)α))

= (p, ταβ(p)v).

Lemma 10.6 now shows that π : E → M is a smooth rank-k vector bundle, with(Uα,Φα) as smooth local trivializations.

Example 171. [Problem 10-7] Compute the transition function for TS2 associatedwith the two local trivializations determined by stereographic coordinates.

See Theorem 128.

Theorem 172. [Problem 10-8] Let Vec1 be the category whose objects are finite-dimensionalreal vector spaces and whose morphisms are linear isomorphisms. If F is a covariantfunctor from Vec1 to itself, for each finite-dimensional vector space V we get a mapF : GL(V ) → GL(F(V )) sending each isomorphism A : V → V to the induced iso-morphism F(A) : F(V ) → F(V ). We say F is a smooth functor if this map issmooth for every V . Given a smooth vector bundle π : E → M and a smooth functor

62

F : Vec1 → Vec1, there is a smooth vector bundle π : F(E) → M whose fiber at eachpoint p ∈M is F(Ep).

Proof. Let k be the rank of π. Define F(E) =∐

p∈M F(Ep) and let π : F(E) → M

be the projection. Let Upp∈M be an open cover of M where each Up contains p and

has an associated local trivialization Φp : π−1(Up) → Up × Rk. Choose an arbitraryisomorphism ϕ : F(Rk) ∼= Rn, where n is the dimension of F(Rk). For each p ∈ M ,

define Φp : π−1(Up)→ Up × Rn as follows: for each x ∈ Up and v ∈ Ex = π−1(x), set

Φp(v) = (x, (ϕ F(Φp|Ex))(v)) where

F(Φp|Ex) : F(Ex)→ F(x × Rk) ∼= F(Rk)

is an isomorphism. For each p, q ∈M with Up ∩ Uq 6= ∅, we have

(Φp Φ−1q )(x, v) = (x, (ϕ F(Φp|Ex) F(Φq|Ex)−1 ϕ−1)(v))

= (x, (ϕ F(Φp|Ex (Φq|Ex)−1) ϕ−1)(v))

= (x, τpq(x)v)

for some smooth τpq since F is a smooth functor. Applying Lemma 10.6 completes theproof.

Theorem 173. [Problem 10-9] Suppose M is a smooth manifold, E →M is a smoothvector bundle, and S ⊆ M is an embedded submanifold with or without boundary. Forany smooth section σ of the restricted bundle E|S → S, there exist a neighborhood U ofS in M and a smooth section σ of E|U such that σ = σ|S. If E has positive rank, thenevery smooth section of E|S extends smoothly to all of M if and only if S is properlyembedded.

Proof. As in Theorem 82.

Theorem 174. [Problem 10-11] Suppose E and E ′ are smooth vector bundles over asmooth manifold M with or without boundary, and F : E → E ′ is a bijective smoothbundle homomorphism over M . Then F is a smooth bundle isomorphism.

Proof. Let k be the rank of E and let k′ be the rank of E ′; let π : E → M andπ′ : E ′ → M be the projections. We want to show that F−1 is smooth. Let v ∈ Eand let Φ : π−1(U)→ U ×Rk be a smooth local trivialization of E satisfying π(v) ∈ Uand πU Φ = π. Let Φ′ : (π′)−1(U ′) → U ′ × Rk′ be a smooth local trivialization of E ′

satisfying π′(F (v)) ∈ U ′ and πU ′ Φ′ = π′. Consider the smooth map f = Φ′ F Φ−1

with its inverse f−1 = Φ F−1 (Φ′)−1 defined on a sufficiently small neighborhood

U ′×Rk′ of Φ′(F (v)). For each x ∈ U the map f |x×Rk is an isomorphism whose inverseis f−1|y(x)×Rk′ , where y(x) = π′(F (ζ(x))) and ζ : M → E is the zero section. Since

63

operator inversion is smooth and y is smooth, f−1 is smooth on U ′×Rk′ . Therefore F−1

is smooth on a neighborhood of F (v), and since v was arbitrary, F−1 is smooth.

Theorem 175. [Problem 10-12] Let π : E →M and π : E →M be two smooth rank-kvector bundles over a smooth manifold M with or without boundary. Suppose Uαα∈Ais an open cover of M such that both E and E admit smooth local trivializations overeach Uα. Let ταβ and ταβ denote the transition functions determined by the given

local trivializations of E and E, respectively. Then E and E are smoothly isomorphicover M if and only if for each α ∈ A there exists a smooth map σα : Uα → GL(k,R)such that

(*) ταβ(p) = σα(p)ταβ(p)σβ(p)−1, p ∈ Uα ∩ Uβ.

Proof. Suppose there is a smooth bundle isomorphism F : E → E. Let α ∈ A and let

Φα : π−1(Uα)→ Uα × Rk and Φα : π−1(Uα)→ Uα × Rk be the associated smooth local

trivializations. Since πUα Φα F = π F = π, the map Φα F : π−1(Uα)→ Uα × Rk

is a smooth local trivialization of E. Therefore

(Φα F Φ−1α )(p, v) = (p, σα(p)v)

for some smooth linear map σα : Uα → GL(k,R). Now let α, β ∈ A. We have

(p, ταβ(p)v) = (Φα Φ−1β )(p, v)

= (Φα F Φ−1α Φα Φ−1

β Φβ F−1 Φ−1β )(p, v)

= (p, σα(p)ταβ(p)σβ(p)−1v),

so ταβ(p) = σα(p)ταβ(p)σβ(p)−1. Conversely, suppose that (*) holds. For each α ∈ A,

define a map Fα : π−1(Uα) → π−1(Uα) by Fα = Φ−1α ϕα Φα, where ϕα(p, v) =

(p, σα(p)v). If α, β ∈ A and Uα ∩ Uβ 6= ∅ then

Fβ = Φ−1β ϕβ Φβ

= Φ−1α Φα Φ−1

β ϕβ Φβ Φ−1α Φα

= Φ−1α ϕα Φα

= Fα

on π−1(Uα) ∩ π−1(Uβ) by (*), so the gluing lemma shows that there is a smooth map

F : E → E such that F |π−1(Uα) = Fα for all α ∈ A. The restriction of F to each fiberis clearly linear. On any π−1(Uα) we have

π F = π Φ−1α ϕα Φα

= πUα ϕα Φα

= πUα Φα

64

= π,

so π F = π on E. By Theorem 174, F is a smooth bundle isomorphism.

Theorem 176. [Problem 10-14] Suppose M is a smooth manifold with or withoutboundary, and S ⊆ M is an immersed submanifold with or without boundary. Iden-tifying TpS as a subspace of TpM for each p ∈ S in the usual way, TS is a smoothsubbundle of TM |S.

Proof. Let ι : S → M be the inclusion map. For each p ∈ S, let Dp = dιp(TpS). LetD =

⋃p∈S Dp = TS ⊆ TM |S. Let p ∈ S, let V be a neighborhood of p in S that is an

embedded submanifold of M , and let (U,ϕ) be a slice chart for V containing p. Writeϕ = (x1, . . . , xn) and let k be the dimension of S. Then V ∩ U is neighborhood of p inS, and the maps σi : V ∩ U → TM |S given by σi(q) = dϕq(∂/∂x

i|q) for i = 1, . . . , ksatisfy the condition in Lemma 10.32.

Theorem 177. [Problem 10-17] Suppose M ⊆ Rn is an immersed submanifold. Theambient tangent bundle TRn|M is isomorphic to the Whitney sum TM ⊕ NM , whereNM →M is the normal bundle.

Proof. Obvious.

Theorem 178. [Problem 10-19] Suppose π : E →M is a fiber bundle with fiber F .

(1) π is an open quotient map.(2) If the bundle is smooth, then π is a smooth submersion.(3) π is a proper map if and only if F is compact.(4) E is compact if and only if both M and F are compact.

Proof. For (1), let V be open in E. Let x ∈ π(V ) and let Φ : π−1(U) → U × F bea local trivialization of E such that x ∈ U . Choose any f ∈ F and a neighborhood

V ⊆ π−1(U) of Φ−1(x, f). Then Φ(V ) is open in U × F since Φ is a homeomorphism,

and πU(Φ(V )) = π(V ) is a neighborhood of x contained in π(V ). This shows thatπ is open, and π is a quotient map since it is surjective. For (2), let v ∈ E and letΦ : π−1(U) → U × F be a smooth local trivialization of E such that π(v) ∈ U . ThenπU Φ = π, so dπv = d(πU)π(v) dΦv is surjective since dΦv and d(πU)π(v) are bothsurjective. This shows that π is a smooth submersion. For (3), suppose that π isproper. If x ∈M then π−1(x) ≈ x × F ≈ F is compact. Conversely, suppose thatF is compact and A ⊆ M is compact. For each p ∈ A, choose a local trivializationΦ : π−1(Up) → Up × F such that p ∈ Up, and choose a precompact neighborhood Vpof p such that Vp ⊆ Up. Since A is compact, the open cover Vp has a finite subcover

65

Vp1 , . . . , Vpk. Then π−1(A) is a closed subset of

k⋃i=1

π−1(Vpi)∼=

k⋃i=1

Vpi × F,

which is compact. Therefore π−1(A) is compact. For (4), if E is compact then M =π(E) is compact and F is compact by (3) since π is proper. Conversely, if M and Fare compact then π is proper by (3), so E = π−1(M) is compact.

Chapter 11. The Cotangent Bundle

Theorem 179. [Exercise 11.10] Suppose M is a smooth manifold and E → M is asmooth vector bundle over M . Define the dual bundle to E to be the bundle E∗ →Mwhose total space is the disjoint union E∗ =

∐p∈M E∗p , where E∗p is the dual space to Ep,

with the obvious projection. Then E∗ →M is a smooth vector bundle, whose transitionfunctions are given by τ ∗(p) = (τ(p)−1)T for any transition function τ : U → GL(k,R)of E.

Proof. Since

τ ∗αβ(p)τ ∗βγ(p) = (ταβ(p)−1)T (τβγ(p)−1)T

= ((ταβ(p)τβγ(p))−1)T

= τ ∗αγ(p),

we can apply Theorem 170.

Theorem 180. [Exercise 11.12] Let M be a smooth manifold with or without boundary,and let ω : M → T ∗M be a rough covector field.

(1) ω is smooth.(2) In every smooth coordinate chart, the component functions of ω are smooth.(3) Each point of M is contained in some coordinate chart in which ω has smooth

component functions.(4) For every smooth vector field X ∈ X(M), the function ω(X) is smooth on M .(5) For every open subset U ⊆M and every smooth vector field X on U , the function

ω(X) : U → R is smooth on U .

Proof. (1) ⇒ (2) ⇒ (3) is clear. If (3) holds for a coordinate chart (U, (xi)) and thenatural coordinates (xi, ξi) on π−1(U) ⊆ T ∗M then the coordinate representation of ωon U is

ω(x) = (x1, . . . , xn, ω1(x), . . . , ωn(x)),

which is smooth if ω1, . . . , ωn are smooth. This proves (3) ⇒ (1). This also implies(3)⇒ (4). Suppose that (4) holds, U ⊆M is open and X is a smooth vector field on U .

66

For each p ∈ U , choose a smooth bump function ψ that is equal to 1 on a neighborhood

of p and supported in U . Define X = ψX, extended to be zero on M \ suppψ. Then

ω(X) is smooth and is equal to ω(X) in a neighborhood of p. So ω(X) is smooth in aneighborhood of every point in U , which proves (4)⇒ (5). If (5) holds then (2) holds,since ωi = ω(∂/∂xi) for any smooth chart (U, (xi)).

Theorem 181. [Exercise 11.16] Let M be a smooth manifold with or without boundary,and let ω be a rough covector field on M . If (εi) is a smooth coframe on an open subsetU ⊆ M , then ω is smooth on U if and only if its component functions with respect to(εi) are smooth.

Proof. This follows from Proposition 10.22.

Example 182. [Exercise 11.17] Let f(x, y) = x2 on R2, and let X be the vector field

X = grad f = 2x∂

∂x.

Compute the coordinate expression for X in polar coordinates (on some open subseton which they are defined) and show that it is not equal to

∂f

∂r

∂

∂r+∂f

∂θ

∂

∂θ.

Set x = r cos θ and y = r sin θ; then f(r, θ) = r2 cos2 θ, so

∂f

∂r

∂

∂r+∂f

∂θ

∂

∂θ= 2r cos2 θ

∂

∂r− 2r2 sin θ cos θ

∂

∂θ.

The derivative of the transition map ϕ(r, θ) = (r cos θ, r sin θ) is[cos θ −r sin θsin θ r cos θ

].

Therefore

X = 2r cos θ

(cos θ

∂

∂r+ sin θ

∂

∂θ

)6= ∂f

∂r

∂

∂r+∂f

∂θ

∂

∂θ.

Theorem 183. [Exercise 11.21] Let M be a smooth manifold with or without boundary,and let f, g ∈ C∞(M).

(1) If a and b are constants, then d(af + bg) = a df + b dg.(2) d(fg) = f dg + g df .(3) d(f/g) = (g df − f dg)/g2 on the set where g 6= 0.(4) If J ⊆ R is an interval containing the image of f , and h : J → R is a smooth

function, then d(h f) = (h′ f) df .(5) If f is constant, then df = 0.

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Proof. For (1), we have

d(af + bg)p(v) = v(af + bg) = avf + bvg = (a dfp + b dgp)(v).

For (2), we have

d(fg)p(v) = v(fg) = f(p)vg + g(p)vf = (f dgp + g dfp)(v).

For (3), if p ∈M and (U, (xi)) is a smooth chart containing p then on U ,

d(f/g)x =n∑i=1

∂(f/g)

∂xi(x) dxi|x

=1

g(x)2

n∑i=1

[g(x)

∂f

∂xi(x)− f(x)

∂g

∂xi(x)

]dxi|x

= (g(x) dfx − f(x) dgx)/g(x)2.

For (4), we have

d(h f)p(v) = dhf(p)(dfp(v)) = h′(f(p))dfp(v).

For (5), if (U, (xi)) is a smooth chart and f is the coordinate representation of f on U

then Df = 0, so f is constant and f is constant on U . The result follows from Theorem29.

Theorem 184. [Exercise 11.24] For a smooth real-valued function f : M → R, p ∈Mis a critical point of f if and only if dfp = 0.

Proof. Obvious.

Theorem 185. [Exercise 11.35] Let M be a smooth manifold with or without boundary.Suppose γ : [a, b]→M is a piecewise smooth curve segment, and ω, ω1, ω2 ∈ X∗(M).

(1) For any c1, c2 ∈ R,ˆγ

(c1ω1 + c2ω2) = c1

ˆγ

ω1 + c2

ˆγ

ω2.

(2) If γ is a constant map, then´γω = 0.

(3) If γ1 = γ|[a,c] and γ2 = γ|[c,b] with a < c < b, thenˆγ

ω =

ˆγ1

ω +

ˆγ2

ω.

(4) If F : M → N is any smooth map and η ∈ X∗(N), thenˆγ

F ∗η =

ˆFγ

η.

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Proof. We can assume that γ is smooth rather than piecewise smooth. For (1),ˆγ

(c1ω1 + c2ω2) =

ˆ[a,b]

γ∗(c1ω1 + c2ω2)

= c1

ˆ[a,b]

γ∗ω1 + c2

ˆ[a,b]

γ∗ω2

= c1

ˆγ

ω1 + c2

ˆγ

ω2

since γ∗ is linear. For (2), if γ is constant then dγ = 0 and γ∗ = 0. For (3),ˆγ

ω =

ˆ[a,b]

γ∗ω

=

ˆ[a,c]

γ∗ω +

ˆ[c,b]

γ∗ω

=

ˆγ1

ω +

ˆγ2

ω.

For (4), ˆγ

F ∗η =

ˆ[a,b]

γ∗F ∗η =

ˆ[a,b]

(F γ)∗η =

ˆFγ

η.

Theorem 186. [Exercise 11.41] A smooth covector field ω is conservative if and onlyif its line integrals are path-independent, in the sense that

´γω =´γω whenever γ and

γ are piecewise smooth curve segments with the same starting and ending points.

Proof. Suppose that ω is conservative, γ is defined on [a, b], and γ is defined on [c, d].Define γ − γ : [a, b+ d− c]→M by

(γ − γ)(t) =

γ(t), 0 ≤ t ≤ b,

γ(b+ d− t), b ≤ t ≤ b+ d− c.

Then γ − γ is piecewise smooth, so

0 =

ˆγ−γ

ω =

ˆγ

ω −ˆγ

ω

by Theorem 185. The converse is obvious since any piecewise smooth closed curvesegment has the same starting and ending points as a constant curve segment.

Theorem 187. [Problem 11-1]

(1) Suppose V and W are finite-dimensional vector spaces and A : V → W is anylinear map. The following diagram commutes:

69

V W

V ∗∗ W ∗∗

A

ξV ξW

(A∗)∗

where ξV and ξW denote the natural isomorphisms.(2) There does not exist a way to assign to each finite-dimensional vector space V

an isomorphism βV : V → V ∗ such that for every linear map A : V → W , thefollowing diagram commutes:

V W

V ∗ W ∗

A

βV βW

A∗

Proof. For (1), we have

((A∗)∗(ξV v))ω = (ξV v)(A∗ω) = (A∗ω)v = ω(Av) = (ξW (Av))ω.

For (2), suppose that there are such isomorphisms βV . Take V = W = R. The diagramcommutes when A = 0, so

βR = 0 βR 0 = 0.

This contradicts the fact that βR is an isomorphism.

Theorem 188. [Problem 11-3] Let Vec1 be the category of finite-dimensional vectorspaces and linear isomorphisms as in Theorem 172. Define a functor F : Vec1 → Vec1

by setting F(V ) = V ∗ for a vector space V , and F(A) = (A−1)∗ for an isomorphismA. Then F is a smooth covariant functor, and for every M , F(TM) and T ∗M arecanonically smoothly isomorphic vector bundles.

Proof. F is covariant because

((A B)−1)∗ = (B−1 A−1)∗ = (A−1)∗ (B−1)∗,

and it is smooth because matrix inversion and transposition are smooth. So Theorem172 shows that F(TM) exists, and we can define a smooth vector bundle isomorphismϕ : F(TM)→ T ∗M by setting ϕ|T ∗pM = IdT ∗pM for each p ∈M .

Theorem 189. [Problem 11-4] Let M be a smooth manifold with or without boundaryand let p be a point of M . Let Jp denote the subspace of C∞(M) consisting of smooth

70

functions that vanish at p, and let J 2p be the subspace of Jp spanned by functions of the

form fg for some f, g ∈ Jp.

(1) f ∈ J 2p if and only if in any smooth local coordinates, its first-order Taylor

polynomial at p is zero.(2) Define a map Φ : Jp → T ∗pM by setting Φ(f) = dfp. The restriction of Φ to J 2

p

is zero, and Φ descends to a vector space isomorphism from Jp/J 2p to T ∗pM .

Proof. If f ∈ J 2p then f =

∑ki=1 αigihi for some αi ∈ R and gi, hi ∈ Jp. If in smooth

local coordinates we have Df1(p) = Df2(p) = 0, then

D(gihi)(p)u = gi(p)[Dhi(p)u] + [Dgi(p)u]hi(p) = 0,

so the first-order Taylor polynomial of f is zero by linearity. Conversely, if Df(p) = 0then Theorem C.15 shows that locally we can write

f(x) =n∑

i,j=1

(xi − pi)(xj − pj)ˆ 1

0

(1− t) ∂2f

∂xi∂xj(p+ t(x− p)) dt,

which is an element of J 2p since xi − pi and xj − pj times the integral are elements of

Jp. Therefore f =∑k

i=1 αigihi for some αi ∈ R and gi, hi ∈ Jp in a neighborhood U

of p. Choose a precompact neighborhood V of p with V ⊆ U and choose a partitionof unity ψ0, ψ1 subordinate to the open cover

U,M \ V

. We take gi = hi = 0 on

M \ suppψ0. Then ψ1, f ∈ Jp and

f =k∑i=1

αi(ψ0gi)hi + ψ1f,

which proves (1). Part (2) follows immediately.

Theorem 190. [Problem 11-5] For any smooth manifold M , T ∗M is a trivial vectorbundle if and only if TM is trivial.

Proof. If TM is trivial then it admits a smooth global frame; Lemma 11.14 shows thatits dual coframe is a smooth global coframe for M , so T ∗M is trivial. The converse issimilar.

Theorem 191. [Problem 11-6] Suppose M is a smooth n-manifold, p ∈ M , andy1, . . . , yk are smooth real-valued functions defined on a neighborhood of p in M .

(1) If k = n and (dy1|p, . . . , dyn|p) is a basis for T ∗pM then (y1, . . . , yn) are smoothcoordinates for M in some neighborhood of p.

(2) If (dy1|p, . . . , dyk|p) is a linearly independent k-tuple of covectors and k < n,then there are smooth functions yk+1, . . . , yn such that (y1, . . . , yn) are smoothcoordinates for M in a neighborhood of p.

71

(3) If (dy1|p, . . . , dyk|p) span T ∗pM , there are indices i1, . . . , in such that (yi1 , . . . , yin)are smooth coordinates for M in a neighborhood of p.

Proof. Part (1) follows from the inverse function theorem. For (2), complete (dy1|p, . . . , dyk|p)to a basis (dy1|p, . . . , dyk|p, ωk+1, . . . , ωn), choose smooth functions yi such that dyi|p =ωi for i = k + 1, . . . , n, and apply (1). For (3), choose indices i1, . . . , in such thatdyi1|p, . . . , dyin|p is a basis for T ∗pM and apply (1).

Example 192. [Problem 11-7] In the following problems, M and N are smooth mani-folds, F : M → N is a smooth map, and ω ∈ X∗(N). Compute F ∗ω in each case.

(1) M = N = R2; F (s, t) = (st, et); ω = x dy − y dx.(2) M = R2 and N = R3; F (θ, ϕ) = ((cosϕ + 2) cos θ, (cosϕ + 2) sin θ, sinϕ);

ω = z2 dx.(3) M = (s, t) ∈ R2 : s2 + t2 < 1 and N = R3 \ 0; F (s, t) = (s, t,

√1− s2 − t2);

ω = (1− x2 − y2) dz.

For (1),

dx = t ds+ s dt, dy = et dt,

so

ω = st(et dt)− et(t ds+ s dt) = −tet ds+ set(t− 1) dt.

For (2),

dx = −(cosϕ+ 2) sin θ dθ − sinϕ cos θ dϕ,

so

ω = sin2 ϕ(−(cosϕ+ 2) sin θ dθ − sinϕ cos θ dϕ).

For (3),

dx = ds,

dy = dt,

dz =−s√

1− s2 − t2ds+

−t√1− s2 − t2

dt,

so

ω = −s√

1− s2 − t2 ds− t√

1− s2 − t2 dt.

Theorem 193. [Problem 11-8]

(1) Suppose F : M → N is a diffeomorphism, and let dF ∗ : T ∗N → T ∗M be themap whose restriction to each cotangent space T ∗qN is equal to dF ∗F−1(q). Then

dF ∗ is a smooth bundle homomorphism.

72

(2) Let Diff1 be the category whose objects are smooth manifolds, but whose onlymorphisms are diffeomorphisms; and let VB be the category whose objects aresmooth vector bundles and whose morphisms are smooth bundle homomorphisms.The assignment M 7→ T ∗M , F 7→ dF ∗ defines a contravariant functor from Diff1

to VB, called the cotangent functor.

Proof. It is clear that dF ∗ is a smooth bundle homomorphism covering F−1. Since

d(F G)∗p = (dFG(p) dGp)∗ = dG∗p dF ∗G(p)

we have d(F G)∗ = dG∗ dF ∗. Also, d(IdM)∗ = IdT ∗M . This shows that M 7→ T ∗M ,F 7→ dF ∗ defines a contravariant functor.

Example 194. [Problem 11-9] Let f : R3 → R be the function f(x, y, z) = x2 +y2 +z2,and let F : R2 → R3 be the following map (the inverse of the stereographic projection):

F (u, v) =

(2u

u2 + v2 + 1,

2v

u2 + v2 + 1,u2 + v2 − 1

u2 + v2 + 1

).

Compute F ∗df and d(f F ) separately, and verify that they are equal.

We havedf = 2x dx+ 2y dy + 2z dz

and

dx =2(u2 + v2 + 1)− 4u2

(u2 + v2 + 1)2du+

−4uv

(u2 + v2 + 1)2dv,

dy =−4uv

(u2 + v2 + 1)2du+

2(u2 + v2 + 1)− 4v2

(u2 + v2 + 1)2dv,

dz =4u

(u2 + v2 + 1)2du+

4v

(u2 + v2 + 1)2dv,

sodf = 0.

Alternatively we have f F = 1, so d(f F ) = 0.

Example 195. [Problem 11-10] In each of the cases below, M is a smooth manifoldand f : M → R is a smooth function. Compute the coordinate representation for df ,and determine the set of all points p ∈M at which dfp = 0.

(1) M = (x, y) ∈ R2 : x > 0; f(x, y) = x/(x2 + y2). Use standard coordinates(x, y).

(2) M and f are as in (1); this time use polar coordinates (r, θ).(3) M = S2 ⊆ R3; f(p) = z(p) (the z-coordinate of p as a point in R3). Use north

and south stereographic coordinates.(4) M = Rn; f(x) = |x|2. Use standard coordinates.

73

For (1),

df =−x2 + y2

(x2 + y2)2dx+

−2xy

(x2 + y2)2dy,

which is zero when x2 = y2 and xy = 0, i.e. never on M . For (2) we have

f(r, θ) =cos θ

r,

so

df = −cos θ

r2dr − sin θ

rdθ,

which is zero when cos θ = sin θ = 0, i.e. never on M . For (3) we have

f(u, v) =u2 + v2 − 1

u2 + v2 + 1

on S2 \N using the north stereographic projection, so

df =4u

(u2 + v2 + 1)2du+

4v

(u2 + v2 + 1)2dv,

which is zero when u = v = 0, i.e. when p is the south pole (0, 0,−1). A similarcalculation using the south stereographic projection shows that dfp = 0 when p is thenorth pole (0, 0, 1). For (4) we have

df = 2x1 dx1 + · · ·+ 2xn dxn,

which is zero when p = 0.

Theorem 196. [Problem 11-11] Let M be a smooth manifold, and C ⊆ M be anembedded submanifold. Let f ∈ C∞(M), and suppose p ∈ C is a point at which fattains a local maximum or minimum value among points in C. Given a smooth localdefining function Φ : U → Rk for C on a neighborhood U of p in M , there are realnumbers λ1, . . . , λk (called Lagrange multipliers) such that

dfp = λ1dΦ1|p + · · ·+ λkdΦk|p.

Proof. Let n be the dimension of M ; then C has dimension n − k. Let (V, ϕ) be a

smooth slice chart for C in M centered at p such that V ⊆ U , let f = f ϕ−1, let

Φ = Φϕ−1, and let p = ϕ(p). It suffices to show that there are real numbers λ1, . . . , λksuch that

Df(p) = λ1DΦ1(p) + · · ·+ λkDΦk(p).

Since Φ−1(0) ⊆ V ∩C, this follows from the method of Lagrange multipliers on Rn.

Theorem 197. [Problem 11-12] Any two points in a connected smooth manifold canbe joined by a smooth curve segment.

74

Proof. Let M be a connected smooth n-manifold. Define an equivalence relation ∼ onM by setting p ∼ q if and only if there is a smooth curve segment from p to q. Thisrelation is clearly reflexive and symmetric; if we can show that it is transitive, then theresult follows from Theorem 29. Suppose p ∼ q and q ∼ r, let γ : [−1, 0] → M be asmooth curve segment from p to q, and let γ′ : [0, 1]→ M be a smooth curve segmentfrom q to r. Let (U,ϕ) be a smooth chart centered at q; by shrinking U , we mayassume that ϕ(U) = Bn. Choose 0 < ε < 1 so that γ((−ε, 0]) ⊆ U and γ′([0, ε)) ⊆ U .Let γ = ϕ γ : (−ε, 0] → U and γ′ = ϕ γ′ : [0, ε) → U . Let ` : (−ε, ε) → Bn begiven by t 7→ (t/ε, 0, . . . , 0). By Proposition 2.25, there is a smooth bump functionψ : (−ε, ε)→ R for [−ε/4, ε/4] supported in (−ε/2, ε/2). Define α : (−ε, ε)→ Bn by

t 7→ ψ(t)`(t) + (1− ψ(t))(γ(t) + γ′(t)),

taking γ = 0 on (0, ε) and γ′ = 0 on (−ε, 0). It is easy to check that α is smooth on(−ε, 0), (−ε/4, ε/4) and (0, ε), so α is smooth. Let α = ϕ−1 α; then α, γ|[−1,ε/2), andγ′|(ε/2,1] agree on their overlaps, so we have a smooth curve segment defined on [−1, 1]from p to r.

Theorem 198. [Problem 11-13] The length of a smooth curve segment γ : [a, b]→ Rn

is defined to be the value of the (ordinary) integral

L(γ) =

ˆ b

a

|γ′(t)| dt.

There is no smooth covector field ω ∈ X∗(Rn) with the property thatˆγ

ω = L(γ)

for every smooth curve γ.

Proof. Let −γ denote the curve t 7→ γ(b− t+ a). Then L(−γ) = L(γ), butˆ−γω = −

ˆγ

ω = −L(γ).

Example 199. [Problem 11-14] Consider the following two covector fields on R3:

ω = − 4z dx

(x2 + 1)2+

2y dy

y2 + 1+

2x dz

x2 + 1,

η = − 4xz dx

(x2 + 1)2+

2y dy

y2 + 1+

2 dz

x2 + 1.

(1) Set up and evaluate the line integral of each covector field along the straightline segment from (0, 0, 0) to (1, 1, 1).

(2) Determine whether either of these covector fields is exact.

75

(3) For each one that is exact, find a potential function and use it to recompute theline integral.

Let γ : [0, 1]→ R3 be given by t 7→ (t, t, t). Thenˆγ

ω =

ˆ 1

0

(− 4t

(t2 + 1)2+

2t

t2 + 1+

2t

t2 + 1

)dt

=

ˆ 1

0

4t3

(1 + t2)2dt

= 2 log 2− 1,

and ˆγ

η =

ˆ 1

0

(− 4t2

(t2 + 1)2+

2t

t2 + 1+

2

t2 + 1

)dt

=

ˆ 1

0

2(t3 − t2 + t+ 1)

(1 + t2)2dt

= log 2 + 1.

It is easy to check that∂

∂zω1 6=

∂

∂xω3,

so ω is not closed and not exact. It is similarly easy to check that η is closed, soTheorem 11.49 shows that η is exact. Suppose f is a potential for η; it must satisfy

∂f

∂x= − 4xz

(x2 + 1)2,

∂f

∂y=

2y

y2 + 1,

∂f

∂z=

2

x2 + 1.

Integrating the first equation, we have

f(x, y, z) =2z

x2 + 1+ C1(y, z).

Then∂C1

∂y=

2y

y2 + 1,

so

C1(y, z) = log(y2 + 1) + C2(z).

Finally we have2

x2 + 1+∂C2

∂z=

2

x2 + 1,

so C2 is constant. It is now easy to check that for any C ∈ R,

f(x, y, z) =2z

x2 + 1+ log(y2 + 1) + C

76

is a potential for η. Thereforeˆγ

η =

ˆγ

df =2

2+ log(2)− 0 = log 2 + 1.

Theorem 200. [Problem 11-15] Let X be a smooth vector field on an open subsetU ⊆ Rn. Given a piecewise smooth curve segment γ : [a, b] → U , define the lineintegral of X over γ, denoted by

´γX · ds, as

ˆγ

X · ds =

ˆ b

a

Xγ(t) · γ′(t) dt,

where the dot on the right-hand side denotes the Euclidean dot product between tangentvectors at γ(t), identified with elements of Rn. A conservative vector field is onewhose line integral around every piecewise smooth closed curve is zero.

(1) X is conservative if and only if there exists a smooth function f ∈ C∞(U) suchthat X = grad f .

(2) Suppose n = 3. If X is conservative then curlX = 0, where

curlX =

(∂X3

∂x2− ∂X2

∂x3

)∂

∂x1+

(∂X1

∂x3− ∂X3

∂x1

)∂

∂x2+

(∂X2

∂x1− ∂X1

∂x2

)∂

∂x3.

(3) If U ⊆ R3 is star-shaped, then X is conservative on U if and only if curlX = 0.

Proof. For any X ∈ X(U), define a smooth covector field ω by ωx(v) = Xx · v; thenˆγ

X · ds =

ˆ b

a

ωγ(t)(γ′(t)) dt =

ˆγ

ω.

Part (1) follows immediately from Theorem 11.42, and part (2) follows from Proposition11.44. Part (3) follows from Theorem 11.49.

Theorem 201. [Problem 11-16] Let M be a compact manifold of positive dimension.Every exact covector field on M vanishes at least at two points in each component ofM .

Proof. Let ω = df be an exact covector field. Let U be a component of M . Since Uis compact, f attains a maximum at some x ∈ U and a minimum at some y ∈ U . Ifx = y then f is constant on U , so df = 0 on U . Otherwise, df vanishes at the distinctpoints x and y.

Theorem 202. [Problem 11-17] Let Tn = S1 × · · · × S1 ⊆ Cn denote the n-torus. Foreach j = 1, . . . , n, let γj : [0, 1]→ Tn be the curve segment

γj(t) = (1, . . . , e2πit, . . . , 1).

A closed covector field ω on Tn is exact if and only if´γjω = 0 for j = 1, . . . , n.

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Proof. Let εn : Rn → Tn be the smooth covering map εn(x1, . . . , xn) = (e2πix1 , . . . , e2πixn).Note that εn is a local diffeomorphism, so Corollary 11.46 shows that (εn)∗ω is closedand Theorem 11.49 shows that (εn)∗ω is exact. If ω is exact then

´γjω = 0 since each

γj is a closed curve segment. Conversely, suppose that´γjω = 0 for j = 1, . . . , n. Let

γ : [0, 1] → Tn be a piecewise smooth closed curve segment and let γ : [0, 1] → Rn bea lift of γ such that γ = εn γ; Theorem 41 shows that γ is smooth. Write γ(0) =(x1, . . . , xn) and γ(1) = (x1+m1, . . . , x

n+mn) for integers m1, . . . ,mn. Assume withoutloss of generality that x1 = · · · = xn = 0. For each i = 1, . . . , n, let αi : [0, 1]→ Rn bethe straight line path from (m1, . . . ,mi−1, 0, . . . , 0) to (m1, . . . ,mi, . . . , 0). Let α be theconcatenation of α1, . . . , αn so that α(0) = 0 and α(1) = (m1, . . . ,mn). Then Theorem186 shows that ˆ

γ

ω =

ˆεnγ

ω

=

ˆγ

(εn)∗ω

=

ˆα

(εn)∗ω

=

ˆα1

(εn)∗ω + · · ·+ˆαn

(εn)∗ω,

=

ˆεnα1

ω + · · ·+ˆεnαn

ω

= 0

since each εn αi is the concatenation of zero or more copies of γi.

Theorem 203. [Problem 11-18] Suppose C and D are categories, and F ,G are (covari-ant or contravariant) functors from C to D. A natural transformation λ from F toG is a rule that assigns to each object X ∈ Ob(C) a morphism λX ∈ HomD(F(X),G(X))in such a way that for every pair of objects X, Y ∈ Ob(C) and every morphism f ∈HomC(X, Y ), the following diagram commutes:

F(X) F(Y )

G(X) G(Y ).

F(f)

λX λY

G(f)

78

(If either F or G is contravariant, the corresponding horizontal arrow should be re-versed.)

(1) Let VecR denote the category of real vector spaces and linear maps, and let D bethe contravariant functor from VecR to itself that sends each vector space to itsdual space and each linear map to its dual map. The assignment V 7→ ξV , whereξV : V → V ∗∗ is the map defined by ξV (v)ω = ω(v), is a natural transformationfrom the identity functor of VecR to D D.

(2) There does not exist a natural transformation from the identity functor of VecRto D.

(3) Let Diff1 be the category of smooth manifolds and diffeomorphisms and VB thecategory of smooth vector bundles and smooth bundle homomorphisms, and letT, T ∗ : Diff1 → VB be the tangent and cotangent functors, respectively. Theredoes not exist a natural transformation from T to T ∗.

(4) Let X : Diff1 → VecR be the covariant functor given by M 7→ X(M), F 7→ F∗;and let X × X : Diff1 → VecR be the covariant functor given by M 7→ X(M) ×X(M), F 7→ F∗ × F∗. The Lie bracket is a natural transformation from X × Xto X.

Proof. Theorem 187 proves (1) and (2). If there is a natural transformation from T toT ∗, then taking the one-point space ∗ produces a natural transformation from theidentity functor of VecR to D, contradicting (2). This proves (3). Corollary 8.31 proves(4).

Chapter 12. Tensors

Theorem 204. [Exercise 12.18] T kT ∗M , T kTM , and T (k,l)TM have natural structuresas smooth vector bundles over M .

Proof. Let n = dimM . Let (U,ϕ) be a smooth chart for M with coordinate functions(xi). Define Φ : π−1(U)→ U × Rn(k+l) by

Φ

(ai1,...,ikj1,...,jl

∂

∂xi1

∣∣∣∣p

⊗ · · · ⊗ ∂

∂xik

∣∣∣∣p

⊗ dxj1∣∣p⊗ · · · ⊗ dxjl

∣∣p

)= (p, (a1,...,1

1,...,1, . . . , an,...,nn...,n )).

Suppose (U , ϕ) is another smooth chart with coordinate functions (xi), and let Φ :

π−1(U)→ U × Rn be defined analogously. It follows from (11.6) and (11.7) that

∂

∂xj

∣∣∣∣p

=∂xi

∂xj(p)

∂

∂xi

∣∣∣∣p

, dxj|p =∂xj

∂xi(p) dxi|p,

79

where x = ϕ ϕ−1 and x = ϕ ϕ−1 = x−1, and p denotes either a point in M or its

coordinate representation as appropriate. Therefore on π−1(U ∩ U) we have

(Φ Φ−1)(p, (a1,...,11,...,1, . . . , a

n,...,nn...,n ))

=

(p,

(∂x1

∂xi1(p) · · · ∂x

1

∂xik(p)

∂xj1

∂x1(p) · · · ∂x

jl

∂x1(p)ai1,...,ikj1,...,jl

,

. . . ,∂xn

∂xi1(p) · · · ∂x

n

∂xik(p)

∂xj1

∂xn(p) · · · ∂x

jl

∂xn(p)ai1,...,ikj1,...,jl

)).

It follows from the vector bundle chart lemma that T ∗M has a smooth structure makingit into a smooth vector bundle for which the maps Φ are smooth local trivializations.

Theorem 205. [Proposition 12.25] Suppose F : M → N and G : N → P are smoothmaps, A and B are covariant tensor fields on N , and f is a real-valued function on N .

(1) F ∗(fB) = (f F )F ∗B.(2) F ∗(A⊗B) = F ∗A⊗ F ∗B.(3) F ∗(A+B) = F ∗A+ F ∗B.(4) F ∗B is a (continuous) tensor field, and is smooth if B is smooth.(5) (G F )∗B = F ∗(G∗B).(6) (IdN)∗B = B.

Proof. For (1),

(F ∗(fB))p = dF ∗p ((fB)F (p)) = dF ∗p (f(F (p))BF (p)) = (f F )(p)(F ∗B)p

since dF ∗p is linear. For (2),

(F ∗(A⊗B))p(v1, . . . , vk+l) = (A⊗B)p(dFp(v1), . . . , dFp(vk+l))

= Ap(dFp(v1), . . . , dFp(vk))Bp(dFp(vk+1), . . . , dFp(vk+l))

= (F ∗A)p(v1, . . . , vk)(F∗B)p(vk+1, . . . , vk+l).

Part (3) follows from the fact that dF ∗p is linear. For (4), let p ∈ M be arbitrary, and

choose smooth coordinates (yj) for N in a neighborhood V of F (p). Let U = F−1(V ),which is a neighborhood of p. Writing B in coordinates as

B = Bj1,...,jkdyj1 ⊗ · · · ⊗ dyjk

for continuous functions Bj1,...,jk on V , we have

F ∗B = F ∗(Bj1,...,jkdyj1 ⊗ · · · ⊗ dyjk)

= (Bj1,...,jk F )F ∗(dyj1 ⊗ · · · ⊗ dyjk)= (Bj1,...,jk F )F ∗dyj1 ⊗ · · · ⊗ F ∗dyjk

= (Bj1,...,jk F )d(yj1 F )⊗ · · · ⊗ d(yjk F ),

80

which is continuous, and is smooth if B is smooth. Part (5) follows from the fact that

d(G F )∗p(α)(v1, . . . , vk) = α(dGF (p)(dFp(v1)), . . . , dGF (p)(dFp(vk)))

= dG∗F (p)(α)(dFp(v1), . . . , dFp(vk))

= dF ∗p (dG∗F (p)(α))(v1, . . . , vk).

Part (6) is obvious.

Example 206. [Problem 12-1] Give an example of finite-dimensional vector spaces Vand W and a specific element α ∈ V ⊗W that cannot be expressed as v⊗w for v ∈ Vand w ∈ W .

Take V = W = R2, let e1 = (1, 0) and e2 = (0, 1), and let α = e1⊗e1 +e2⊗e2. Supposeα = v ⊗ w for v, w ∈ R2, and write v = viei and w = wjej. Then α = viwjei ⊗ ej, so

v1w1 = 1, v1w2 = 0, v2w1 = 0, v2w2 = 1.

This implies that w1 6= 0, so v2 = 0, contradicting v2w2 = 1.

Theorem 207. [Problem 12-2] For any finite-dimensional real vector space V , thereare canonical isomorphisms R⊗ V ∼= V ∼= V ⊗ R.

Proof. We can define a multilinear map f : R×V → V by (r, v) 7→ rv, which induces a

linear map f : R⊗ V → V satisfying f(r ⊗ v) = rv. It is easy to check that the linear

map v 7→ 1 ⊗ v is an inverse to f , so f is an isomorphism. A similar argument showsthat V ∼= V ⊗ R.

Theorem 208. [Problem 12-3] Let V and W be finite-dimensional real vector spaces.The tensor product space V ⊗W is uniquely determined up to canonical isomorphismby its characteristic property. More precisely, suppose π : V ×W → Z is a bilinear mapinto a vector space Z with the following property: for any bilinear map B : V ×W → Y ,

there is a unique linear map B : Z → Y such that the following diagram commutes:

V ×W Y.

Z

B

πB

Then there is a unique isomorphism Φ : V ⊗ W → Z such that π = Φ π, whereπ : V ×W → V ⊗W is the canonical projection.

Proof. Since π is bilinear, there exists a unique linear map Φ : V ⊗W → Z such thatπ = Φ π. Similarly, there exists a unique linear map Ψ : Z → V ⊗ W such that

81

π = Ψ π. We have Φ Ψ π = Φ π = π, so Φ Ψ = IdZ by uniqueness. Similarly,Ψ Φ π = Ψ π = π implies that Ψ Φ = IdV⊗W by uniqueness. Therefore Φ is anisomorphism.

Theorem 209. [Problem 12-4] Let V1, . . . , Vk and W be finite-dimensional real vectorspaces. There is a canonical isomorphism

V ∗1 ⊗ · · · ⊗ V ∗k ⊗W ∼= L(V1, . . . , Vk;W ).

Proof. Proposition 12.10 shows that

V ∗1 ⊗ · · · ⊗ V ∗k ∼= L(V1, . . . , Vk;R),

so it remains to show that

L(V1, . . . , Vk;R)⊗W ∼= L(V1, . . . , Vk;W ).

The bilinear map ϕ : L(V1, . . . , Vk;R)×W → L(V1, . . . , Vk;W ) defined by

ϕ(f, w)(v1, . . . , vk) = f(v1, . . . , vk)w

induces a linear map ϕ : L(V1, . . . , Vk;R)⊗W → L(V1, . . . , Vk;W ) satisfying

ϕ(f ⊗ w)(v1, . . . , vk) = f(v1, . . . , vk)w.

It is easy to check that ϕ takes a basis of L(V1, . . . , Vk;R)⊗W to a basis of L(V1, . . . , Vk;W ),so ϕ is an isomorphism.

Theorem 210. [Problem 12-5] Let V be an n-dimensional real vector space. Then

dim Σk(V ∗) =

(n+ k − 1

k

).

Proof. Let ε1, . . . , εn be a basis for V ∗; we want to show that

B =εi1 · · · εik : 1 ≤ i1 ≤ · · · ≤ ik ≤ n

is a basis for Σk(V ∗), from which the result follows immediately. It is clear that B islinearly independent. Let α ∈ Σk(V ∗) and write

α = ai1,...,ikεi1 ⊗ · · · ⊗ εik .

We have

α = Symα = ai1,...,ik Sym εi1 ⊗ · · · ⊗ εik

= ai1,...,ikεi1 · · · εik ,

which is in the span of B after reordering the indices i1, . . . , ik in each term εi1 · · · εikto be in ascending order. This shows that B is a basis for Σk(V ∗).

Theorem 211. [Problem 12-6]

82

(1) Let α be a covariant k-tensor on a finite-dimensional real vector space V . ThenSymα is the unique symmetric k-tensor satisfying

(Symα)(v, . . . , v) = α(v, . . . , v)

for all v ∈ V .(2) The symmetric product is associative: for all symmetric tensors α, β, γ,

(αβ)γ = α(βγ).

(3) Let ω1, . . . , ωk be covectors on a finite-dimensional vector space. Their symmet-ric product satisfies

ω1 · · ·ωk =1

k!

∑σ∈Sk

ωσ(1) ⊗ · · · ⊗ ωσ(k).

Proof. Let E1, . . . , Ek be a basis for V . For (1),

(Symα)(v, . . . , v) =1

k!

∑σ∈Sk

α(v, . . . , v) = α(v, . . . , v).

The uniqueness of Symα follows from the fact that(Ei1 + · · ·+ Ein)⊗k : 1 ≤ n ≤ k, 1 ≤ i1 < · · · < in ≤ k

spans V ⊗k. For (2), we have

(Sym(Symα⊗ β)⊗ γ)(v, . . . , v) = (Symα⊗ β)(v, . . . , v)γ(v, . . . , v)

= α(v, . . . , v)β(v, . . . , v)γ(v, . . . , v)

= α(v, . . . , v)(Sym β ⊗ γ)(v, . . . , v)

= (Symα⊗ (Sym β ⊗ γ))(v, . . . , v),

so the result follows from (1). (Note that the tuples (v, . . . , v) may have differentnumbers of elements.) Part (3) is obvious.

Example 212. [Problem 12-7] Let (e1, e2, e3) be the standard dual basis for (R3)∗. Thetensor e1⊗e2⊗e3 is not equal to a sum of an alternating tensor and a symmetric tensor.

Let (e1, e2, e3) be the standard basis for R3. Suppose that

e1 ⊗ e2 ⊗ e3 = α + β

where α is alternating and β is symmetric. We have

1 = (e1 ⊗ e2 ⊗ e3)(e1, e2, e3) = α(e1, e2, e3) + β(e1, e2, e3)

but

0 = (e1 ⊗ e2 ⊗ e3)(e2, e3, e1) = α(e2, e3, e1) + β(e2, e3, e1)

= α(e1, e2, e3) + β(e1, e2, e3),

83

which is a contradiction.

Chapter 13. Riemannian Metrics

Theorem 213. [Exercise 13.21] Let (M, g) be a Riemannian n-manifold with or with-out boundary. For any immersed k-dimensional submanifold S ⊆ M with or withoutboundary, the normal bundle NS is a smooth rank-(n − k) subbundle of TM |S. Foreach p ∈ S, there is a smooth frame for NS on a neighborhood of p that is orthonormalwith respect to g.

Proof. Let p ∈ M be arbitrary, and let (X1, . . . , Xk) be a smooth local frame for TSover some neighborhood V of p in M . Because immersed submanifolds are locallyembedded, by shrinking V if necessary, we may assume that it is a single slice in somecoordinate ball or half-ball U ⊆ Rn. Since V is closed in U , Theorem 122(c) shows

that we can complete (X1, . . . , Xk) to a smooth local frame (X1, . . . , Xn) for TM |S overU , and then Proposition 13.6 yields a smooth orthonormal frame (Ej) over U suchthat span(E1|p, . . . , Ek|p) = span(X1|p, . . . , Xk|p) = TpS for each p ∈ U . It followsthat (Ek+1, . . . , En) restricts to a smooth orthonormal frame for NS over V . Thus NSsatisfies the local frame criterion in Lemma 10.32, and is therefore a smooth subbundleof TM |S.

Theorem 214. [Exercise 13.24] Suppose (M, g) and (M, g) are Riemannian manifolds

with or without boundary, and F : M → M is a local isometry. Then Lg(F γ) = Lg(γ)for every piecewise smooth curve segment γ in M .

Proof. We have

Lg(F γ) =

ˆ b

a

|(F γ)′(t)|g dt

=

ˆ b

a

∣∣dFγ(t)(γ′(t))

∣∣gdt

=

ˆ b

a

√g(Fγ)(t)(dFγ(t)(γ′(t)), dFγ(t)(γ′(t))) dt

=

ˆ b

a

√(F ∗g)(γ′(t), γ′(t)) dt

=

ˆ b

a

√g(γ′(t), γ′(t)) dt

= Lg(γ).

84

Theorem 215. [Exercise 13.27] Suppose (M, g) and (M, g) are connected Riemannian

manifolds and F : M → M is a Riemannian isometry. Then

dg(F (p), F (q)) = dg(p, q)

for all p, q ∈M .

Proof. If γ is a piecewise smooth curve segment from p to q, then F γ is a piecewisesmooth curve segment from F (p) to F (q), and Theorem 214 shows that Lg(γ) = Lg(F γ). Similarly, if γ is a piecewise smooth curve segment from F (p) to F (q) then F−1 γis a piecewise smooth curve segment from p to q and Lg(γ) = Lg(F

−1 γ).

Theorem 216. [Problem 13-1] If (M, g) is a Riemannian n-manifold with or without

boundary, let UM ⊆ TM be the subset UM =

(x, v) ∈ TM : |v|g = 1

, called the

unit tangent bundle of M . Then UM is a smooth fiber bundle over M with modelfiber Sn−1.

Proof. Define ` : TM → R by (x, v) 7→ g(x)(v, v) and for fixed x ∈ M , define`x : TxM → R by v 7→ g(x)(v, v). In any smooth local coordinates around x wehave D`(v)(u) = 2g(x)(v, u) and in particular D`(v)(v) = 2g(x)(v, v) > 0 for all v 6= 0.Therefore Corollary 5.14 shows that UM = `−1(1) is a properly embedded submani-fold of TM . Let π : TM → M be the canonical projection. Let Φ : π−1(U)→ U × Rn

be a smooth local trivialization. Define

ϕ : π−1(U) ∩ UM → U × Sn−1

(x, v) 7→(x,

(πRn Φ)(x, v)

‖(πRn Φ)(x, v)‖

)and

ψ : U × Sn−1 → π−1(U) ∩ UM

(x, v) 7→ Φ−1(x, v)

|Φ−1(x, v)|g.

Then ϕ and ψ are smooth and inverses of each other, so ϕ is a local trivialization ofUM over M .

Theorem 217. [Problem 13-2] Suppose that E is a smooth vector bundle over a smoothmanifold M with or without boundary, and V ⊆ E is an open subset with the propertythat for each p ∈ M , the intersection of V with the fiber Ep is convex and nonempty.By a “section of V ”, we mean a (local or global) section of E whose image lies in V .

(1) There exists a smooth global section of V .

85

(2) Suppose σ : A→ V is a smooth section of V defined on a closed subset A ⊆M .(This means that σ extends to a smooth section of V in a neighborhood of eachpoint of A.) There exists a smooth global section σ of V whose restriction to Ais equal to σ. If V contains the image of the zero section of E, then σ can bechosen to be supported in any predetermined neighborhood of A.

Proof. For (1), let p ∈ M and choose a local trivialization Φ : π−1(U) → U × Rn suchthat p ∈ U . Choose any v ∈ π−1(p) and choose neighborhoods Up of p and Fp of v′ =(πRn Φ)(v) such that Up×Fp ⊆ Φ(π−1(U)∩V ). Define τp : Up → V by x 7→ Φ−1(x, v′).The family of sets Up : p ∈M is an open cover of M . Let ψp : p ∈M be a smoothpartition of unity subordinate to this cover, with suppψp ⊆ Up. For each p ∈ M , theproduct ψpτp is smooth on Wp by Theorem 159, and has a smooth extension to all of Mif we interpret it to be zero on M \ suppψp. (The extended function is smooth becausethe two definitions agree on the open subset Wp \ suppψp where they overlap.) Thuswe can define τ : M → V by

τ(x) =∑p∈M

ψp(x)τp(x).

Because the collection of supports suppψp is locally finite, this sum actually has onlya finite number of nonzero terms in a neighborhood of any point of M , and thereforedefines a smooth function. The fact that

∑p∈M ψp = 1 implies that τ(x) is a convex

combination of vectors in Ex for every x ∈M , so the image of τ does indeed lie in V .

For (2), let τ : M → V be a smooth global section of V . If V contains the imaage ofthe zero section of E, then let U be some neighborhood of A and replace τ with thezero section of E. For each p ∈ A, choose a neighborhood Wp of p and a local sectionσp : Wp → V of V that agrees with σ on Wp∩A. If U was chosen, then replacing Wp byWp ∩U we may assume that Wp ⊆ U . The family of sets Wp : p ∈ A∪M \ A is anopen cover of M . Let ψp : p ∈ A ∪ ψ0 be a smooth partition of unity subordinateto this cover, with suppψp ⊆ Wp and suppψ0 ⊆M \ A.

For each p ∈ A, the product ψpσp is smooth on Wp and has a smooth extension to allof M if we interpret it to be zero on M \ suppψp. Thus we can define σ : M → V by

σ(x) = ψ0(x)τ(x) +∑p∈A

ψp(x)σp(x).

If x ∈ A, then ψ0(x) = 0 and σp(x) = σ(x) for each p such that ψp(x) 6= 0, so

σ(x) =∑p∈A

ψp(x)σ(x) =

(ψ0(x) +

∑p∈A

ψp(x)

)σ(x) = σ(x),

so σ is indeed an extension of σ. The fact that ψ0 +∑

p∈A ψp = 1 implies that σ(x) isa convex combination of vectors in Ex for every x ∈M , so the image of σ does indeed

86

lie in V . If U was chosen, then it follows from Lemma 1.13(b) that

supp σ =⋃p∈A

suppψp =⋃p∈A

suppψp ⊆ U.

Theorem 218. Suppose (M, g) is a Riemannian manifold, and X is a vector field onM . Given any positive continuous function δ : M → R, there exists a smooth vector

field X ∈ X(M) that is δ-close to X. If X is smooth on a closed subset A ⊆ M ,

then X can be chosen to be equal to X on A. (Two vector fields X, X are δ-close if∣∣∣Xx − Xx

∣∣∣g< δ(x) for all x ∈M .

Proof. If X is smooth on the closed subset A, then by Lemma 8.6, there is a smoothvector field X0 ∈ X(M) that agrees with X on A. Let

U0 = y ∈M : |X0|y −Xy|g < δ(y).

Then U0 is an open subset containing A. (If there is no such set A, we just takeU0 = A = ∅ and F0 = 0.)

We will show that there are countably many points xi∞i=1 in M \ A and smoothcoordinate balls Ui of xi in M \ A such that Ui∞i=1 is an open cover of M \ A and

(*) |Xy −Xi|y|g < δ(y)

for all y ∈ Ui, where Xi ∈ X(Ui) is a constant coefficient vector field such that Xi|xi =Xxi . To see this, for any x ∈ M \ A, let Ux be a smooth coordinate ball around xcontained in M \ A and small enough that

δ(y) >1

2δ(x) and

∣∣Xy −X(x)|y∣∣g<

1

2δ(x)

for all y ∈ Ux, where X(x) ∈ X(Ux) is a constant coefficient vector field such thatX(x)|x = Xx. (Such a neighborhood exists by continuity of δ and X.) Then if y ∈ Ux,we have ∣∣Xy −X(x)|y

∣∣g<

1

2δ(x) < δ(y).

The collection Ux : x ∈M \ A is an open cover of M \ A. Choosing a countablesubcover Uxi

∞i=1 and setting Ui = Uxi and Xi = X(xi), we have (*).

Let ϕ0, ϕi be a smooth partition of unity subordinate to the cover U0, Ui of M , and

define X ∈ X(M) by

Xy = ϕ0(y)X0|y +∑i≥1

ϕi(y)Xi|y.

87

Then clearly X is smooth, and is equal to X on A. For any y ∈ M , the fact that∑i≥0 ϕi = 1 implies that∣∣∣Xy −Xy

∣∣∣g

=

∣∣∣∣∣ϕ0(y)X0|y +∑i≥1

ϕi(y)Xi|y −

(ϕ0(y) +

∑i≥1

ϕi(y)

)Xy

∣∣∣∣∣g

≤ ϕ0(y) |X0|y −Xy|g +∑i≥1

ϕi(y) |Xi|y −Xy|g

< ϕ0(y)δ(y) +∑i≥1

ϕi(y)δ(y) = δ(y).

Lemma 219. Let V be a real inner product space and let ‖·‖ be the norm on V . Letu1, . . . , uk be an orthonormal set of vectors in V with ‖ui‖ = 1 for i = 1, . . . , k. Letv ∈ V with ‖v‖ = 1.

(1) We have

〈u1, v〉2 + · · ·+ 〈uk, v〉2 ≤ 1,

and equality holds if and only if the set v, u1, . . . , uk is linearly dependent.(2) If w ∈ V and

‖v − w‖2 < 1− 〈u1, v〉2 − · · · − 〈uk, v〉2 ,

then w, u1, . . . , uk are linearly independent.

Proof. For (1), complete u1, . . . , uk to an orthonormal basis u1, . . . , un of V . Writev = a1u1 + · · ·+ anun where a1, . . . , an ∈ R. Then

〈u1, v〉2 + · · ·+ 〈uk, v〉2 = 〈u1, a1u1〉2 + · · ·+ 〈uk, akuk〉2

= a21 + · · ·+ a2

k

= 1− a2k+1 − · · · − a2

n

since ‖v‖ = 1. Clearly the last expression is equal to 1 if and only if ak+1, . . . , an = 0,which is true if and only if v is in the span of u1, . . . , uk. For (2), suppose w =a1u1 + · · ·+ akuk where a1, . . . , ak ∈ R. Then

0 >

∥∥∥∥∥v −k∑i=1

aiui

∥∥∥∥∥2

− 1 +k∑i=1

〈ui, v〉2

= 1− 2

⟨v,

k∑i=1

aiui

⟩+

∥∥∥∥∥k∑i=1

aiui

∥∥∥∥∥2

− 1 +k∑i=1

〈ui, v〉2

88

=k∑i=1

(a2i − 2ai 〈ui, v〉+ 〈ui, v〉2)

=k∑i=1

(ai − 〈ui, v〉)2,

which is a contradiction.

Theorem 220. [Problem 13-3] Let M be a smooth manifold.

(1) If there exists a global nonvanishing vector field on M , then there exists a globalsmooth nonvanishing vector field.

(2) If there exists a linearly independent k-tuple of vector fields on M , then thereexists such a k-tuple of smooth vector fields.

Proof. Let g be a Riemannian metric on M . For (1), let X be a global nonvanishing

vector field on M . By Theorem 218, there is a smooth vector field X ∈ X(M) suchthat

|Xx − Xx|g < |Xx|gfor all x ∈ M . This implies that |Xx|g > 0 for all x ∈ M , so X is nonvanishing. For(2), let (X1, . . . , Xk) be a linearly independent k-tuple of vector fields on M . Using theGram-Schmidt process, we may assume that (X1, . . . , Xk) is an orthonormal k-tuple ofvector fields. Assume that X1, . . . , Xj−1 are already smooth. Define

δ(x) =

√1−

∑i 6=j

〈Xj|x, Xi|x〉2g;

then δ : M → R is a positive continuous function by Lemma 219. By Theorem 218,

there is a smooth vector field Xj ∈ X(M) such that

|(Xj|x − Xj|x)|2g < 1−∑i 6=j

〈Xj|x, Xi|x〉2g

for all x ∈ M . Then Lemma 219 shows that (X1, . . . , Xj, . . . , Xk) is linearly inde-

pendent. Apply the Gram-Schmidt process to (X1, . . . , Xj) to obtain an orthonormalk-tuple of vector fields such that the first j vector fields are smooth. Repeating thisprocedure k times produces an orthonormal k-tuple of smooth vector fields.

Example 221. [Problem 13-4] Letg denote the round metric on Sn. Compute the

coordinate representation ofg in stereographic coordinates.

We have

(x1, . . . , xn+1) =(2u1, . . . , 2un, |u|2 − 1)

|u|2 + 1,

89

so

dxj =2(|u|2 − 2uj + 1)

(|u|2 + 1)2duj −

∑i 6=j

4uiuj

(|u|2 + 1)2dui

for each j = 1, . . . , n and

dxn+1 =n∑i=1

4ui

(|u|2 + 1)2dui.

Thereforeg = (dx1)2 + · · ·+ (dxn+1)2

=4

(|u|2 + 1)2((du1)2 + · · ·+ (dun)2).

Theorem 222. [Problem 13-5] Suppose (M, g) is a Riemannian manifold. A smoothcurve γ : J →M is said to be a unit-speed curve if |γ′(t)|g = 1. Every smooth curvewith nowhere-vanishing velocity has a unit-speed reparametrization.

Proof. We can assume that J is open. Choose any t0 ∈ J and define ` : J → R by

`(t) =

ˆ t

t0

|γ′(x)|g dx.

Then x 7→ |γ′(x)|g is smooth since v 7→ |v|g is smooth on (x, v) ∈ TM : v 6= 0 and

γ′(x) 6= 0 for all x ∈ J . Therefore ` is smooth, and since `′(t) 6= 0 for all t ∈ J , theinverse function theorem shows that `−1 : `(J) → J is smooth. Define γ1 : `(J) → Mby γ1 = γ `−1. We have

γ′1(x) = (`−1)′(x)γ′(`−1(x))

=γ′(`−1(x))

`′(`−1(x))

=γ′(`−1(x))

|γ′(`−1(x))|g,

so γ1 is a unit-speed reparametrization of γ.

Theorem 223. [Problem 13-6] Every Riemannian 1-manifold is flat.

Proof. Let (M, g) be a Riemannian 1-manifold. Suppose x ∈M and (U,ϕ) is a smoothchart containing x. By shrinking U , we can assume that U is connected. Therefore ϕ−1 :ϕ(U) → U is a smooth curve, and Theorem 222 shows that there is a diffeomorphismψ : E → ϕ(U) such that ϕ−1 ψ is a unit-speed curve. We want to show that ϕ−1 ψ :

90

E → U is an isometry. If v, w ∈ TpE are nonzero vectors then v = r d/dt|p and

w = s d/dt|p for some r, s ∈ R, so

((ϕ−1 ψ)∗gp)(v, w) = rs((ϕ−1 ψ)∗gp)

(d

dt

∣∣∣∣p

,d

dt

∣∣∣∣p

)

= rs

∣∣∣∣∣d(ϕ−1 ψ)p

(d

dt

∣∣∣∣p

)∣∣∣∣∣2

g

= rs∣∣(ϕ−1 ψ)′(p)

∣∣2g

= rs

= g(ϕ−1ψ)(p)(v, w).

Theorem 224. [Problem 13-7] A product of flat metrics is flat.

Proof. If (R1, g1), . . . , (Rk, gk) are Riemannian manifolds, then

(R1 × · · · ×Rk, g1 ⊕ · · · ⊕ gk)is also a Riemannian manifold. Let (x1, . . . , xk) ∈ R1× · · · ×Rk. For each i = 1, . . . , k,let Fi : Ui → Vi be an isometry such that xi ∈ Ui. Then F1 × · · · × Fk is an isometryfrom the neighborhood U1 × · · · × Uk to the open set V1 × · · · × Vk, since

(F1 × · · · × Fk)∗(g ⊕ · · · ⊕ g) = F ∗1 g ⊕ · · · ⊕ F ∗k g= g1 ⊕ · · · ⊕ gk.

Corollary 225. [Problem 13-8] Let Tn = S1 × · · · × S1 ⊆ Cn, and let g be the metricon Tn induced from the Euclidean metric on Cn (identified with R2n). Then g is flat.

Theorem 226. [Problem 13-10] The shortest path between two points in Euclideanspace is a straight line segment. More precisely, for x, y ∈ Rn, let γ : [0, 1]→ Rn be thecurve segment γ(t) = x + t(y − x). Any other piecewise smooth curve segment γ fromx to y satisfies Lg(γ) > Lg(γ) unless γ is a reparametrization of γ.

Proof. First note that if γ : [a, b]→ Rn thenˆ b

a

|γ′(t)| dt ≥∣∣∣∣ˆ b

a

γ′(t)dt

∣∣∣∣ = |γ(b)− γ(a)| .

By rotating Rn, we can assume that x, y ∈ X, where X = R × 0n−1 is the x1-axis. Let γ : [a, b] → Rn be a piecewise smooth curve segment from x to y such that

91

Lg(γ) = |y − x|. Suppose that γ(c) /∈ X for some c ∈ (a, b). Then

Lg(γ) =

ˆ c

a

|γ′(t)| dt+

ˆ b

c

|γ′(t)| dt

≥ |γ(c)− γ(a)|+ |γ(b)− γ(c)|> |y − x| = Lg(γ),

which is a contradiction. Therefore γ([a, b]) ⊆ X. Let π : Rn → X be the canonicalprojection. A similar argument shows that π γ is a diffeomorphism from [a, b] to theline segment from π(x) to π(y).

Example 227. [Problem 13-11] Let M = R2 \ 0, and let g be the restriction to Mof the Euclidean metric g. Show that there are points p, q ∈ M for which there is nopiecewise smooth curve segment γ from p to q in M with Lg(γ) = dg(p, q).

Let p = (−1, 0) and q = (1, 0). For each 0 < r < 1, let

p′ = (−r, 0), q′ = (r, 0),

p′′ = (−r, r), q′′ = (r, r)

and define αr : [0, 5]→M by

αr(t) =

(1− t)p+ tp′, 0 ≤ t < 1,

(2− t)p′ + (t− 1)p′′, 1 ≤ t < 2,

(3− t)p′′ + (t− 2)q′′, 2 ≤ t < 3,

(4− t)q′′ + (t− 3)q′, 3 ≤ t < 4,

(5− t)q′ + (t− 4)q, 4 ≤ t ≤ 5.

We have

Lg(αr) = 1− r + 4r + 1− r = 2 + 2r

→ 2

as r → 0. Therefore dg(p, q) ≤ 2. But Theorem 226 shows that dg(p, q) ≥ 2, sodg(p, q) = 2. There is no piecewise smooth curve segment γ from p to q such thatLg(γ) = 2, due to Theorem 226.

Theorem 228. [Problem 13-12] Consider Rn as a Riemannian manifold with the Eu-clidean metric g.

(1) Suppose U, V ⊆ Rn are connected open sets, ϕ, ψ : U → V are Riemannianisometries, and for some p ∈ U they satisfy ϕ(p) = ψ(p) and dϕp = dψp. Thenϕ = ψ.

(2) The set of maps from Rn to itself given by the action of E(n) on Rn describedin Example 7.32 is the full group of Riemannian isometries of (Rn, g).

92

Proof. By considering ϕ ψ−1 : V → V , it suffices to show that for any connectedopen set U ⊆ Rn and any Riemannian isometry f : U → U satisfying f(p) = p anddfp = IdTpU for some p ∈ U , we have f = IdU . If γ : I → U is a smooth curvethen Lg(γ) = Lg(f γ), so Theorem 226 implies that f takes straight line segments tostraight line segments (of the same length). Let

E = x ∈ U : f(x) = x, dfx = IdTxU ;

then E is nonempty. Let x ∈ E and let B = Br(x) be an open ball of radius r aroundx such that B ⊆ U . For any v ∈ Sn−1, let γ : [−1, 1] → U be the curve t 7→ x + trvand let L = γ([−1, 1]). Then f γ is a straight line segment of length 2r such that(f γ)(0) = x. But d(f γ)0 = dfx dγ0 = dγ0, so (f γ)([−1, 1]) = L. Using the factthat

dg(x, y) = dg(f(x), f(y)) = dg(x, f(y))

for all y ∈ U , we conclude that f γ = γ. But v was arbitrary, so f |B is the identity mapon B(x). This shows that B ⊆ E, and therefore E is open. It is clear from continuitythat E is closed. Since U is connected, we must have E = U . This proves (1).

Let ϕ : Rn → Rn be a Riemannian isometry. Then ϕ(x) = Dϕ(0)−1(ϕ(x) − ϕ(0))defines a Riemannian isometry satisfying Dϕ(0) = I and ϕ(0) = 0, so (1) shows thatϕ = IdRn . Therefore ϕ(x) = ϕ(0) +Dϕ(0)x is an affine transformation.

Theorem 229. [Problem 13-15] Let (M, g) be a Riemannian manifold, and let g bethe product metric on M × R determined by g and the Euclidean metric on R. LetX = 0⊕ d/dt be the product vector field on M × R determined by the zero vector fieldon M and the standard coordinate vector field d/dt on R. Then X is a Killing vectorfield for (M × R, g).

Proof. The flow of X is given by θ((x, r), t) = (x, r + t).

Theorem 230. [Problem 13-16] If g = f(t)dt2 is a Riemannian metric on R, then gis complete if and only if both of the following improper integrals diverge:

(*)

ˆ ∞0

√f(t) dt,

ˆ 0

−∞

√f(t) dt.

Proof. If x, y ∈ R and γ : [x, y]→ R is the identity map then

dg(x, y) =

ˆ y

x

|γ′(t)|g dt =

ˆ y

x

|1|g dt =

ˆ y

x

√f(t) dt.

Note that f(t) > 0 for all t ∈ R. If´∞

0

√f(t) dt converges, then xn = 1, 2, . . .

is a Cauchy sequence. Suppose that g is complete; then n → x for some x ∈ R.If x ≥ 0 then dg(n, x) =

´ nx

√f(t) dt converges to a positive number as n → ∞,

which is a contradiction. A similar contradiction arises if x < 0. Therefore g is not

93

complete. Conversely, suppose that both integrals in (*) diverge and let xn be aCauchy sequence. Then xn is contained in a set K ⊆ R that is bounded with respectto the metric induced by g. Since the integrals in (*) diverge, K must also be boundedwith respect to the Euclidean metric. By replacing K with a closed and boundedinterval containing K, we may assume that K is compact. There are positive constantsc, C such that

c ≤√f(t) ≤ C and

√f(t)/C ≤ 1 ≤

√f(t)/c

for all t ∈ K. Let ε > 0. Since xn is Cauchy, there exists an integer N such that

−cε <ˆ xn

xm

√f(t) dt < cε

for all m,n ≥ N . Then

−ε < −cεC<

1

C

ˆ xn

xm

√f(t) dt ≤ |xn − xm| ≤

1

c

ˆ xn

xm

√f(t) dt <

cε

c= ε,

so in the Euclidean metric, xn is Cauchy and xn → x for some x ∈ K. We have∣∣∣∣ˆ x

xn

√f(t) dt

∣∣∣∣→ 0

as n→∞, so xn → x in the metric induced by g.

Theorem 231. [Problem 13-18] Suppose (M, g) is a connected Riemannian manifold,S ⊆M is a connected embedded submanifold, and g is the induced Riemannian metricon S.

(1) dg(p, q) ≥ dg(p, q) for p, q ∈ S.(2) If (M, g) is complete and S is properly embedded, then (S, g) is complete.(3) Every connected smooth manifold admits a complete Riemannian metric.

Proof. Let ι : S → M be the inclusion map. If γ : [a, b] → S is a piecewise smoothcurve segment from p to q in S then ι γ is a piecewise smooth curve segment fromp to q in M and Lg(γ) = Lg(ι γ) ≥ dg(p, q). Since this holds for all γ, we havedg(p, q) ≥ dg(p, q). This proves (1). Let xn be a Cauchy sequence in S and let ε > 0.There exists an integer N such that

dg(xm, xn) ≤ dg(xm, xn) < ε

for all m,n ≥ N , so xn is a Cauchy sequence in M and xn → x for some x ∈M . ButS is closed in M , so x ∈ S. This proves (2). Let N be a connected smooth manifold.Theorem 6.15 shows that there is a proper smooth embedding j : N → R2n+1. Since(R2n+1, g) is complete, (N, j∗g) is complete by (2). This proves (3).

94

Example 232. [Problem 13-19] The following example shows that the converse ofTheorem 231(b) does not hold. Define F : R→ R2 by F (t) = ((et+1) cos t, (et+1) sin t).Show that F is an embedding that is not proper, yet R is complete in the metric inducedfrom the Euclidean metric on R2.

We have

DF (t) =

[et(cos t− sin t)− sin tet(cos t+ sin t) + cos t

],

which is never zero. Since F is an open map, it is a smooth embedding. But F−1(B2(0)

)is not compact, where B2(0) is the closed ball of radius 2 around (0, 0). The metric onR induced from the Euclidean metric on R2 is given by

f ∗g = d((et + 1) cos t)2 + d((et + 1) sin t)2

= (2e2t + 2et + 1)dt2.

But ˆ ∞0

√2e2t + 2et + 1 dt and

ˆ 0

−∞

√2e2t + 2et + 1 dt

diverge, so Theorem 230 shows that f ∗g is complete.

Theorem 233. [Problem 13-21] Let (M, g) be a Riemannian manifold, let f ∈ C∞(M),and let p ∈M be a regular point of f .

(1) Among all unit vectors v ∈ TpM , the directional derivative vf is greatest whenv points in the same direction as grad f |p, and the length of grad f |p is equal tothe value of the directional derivative in that direction.

(2) grad f |p is normal to the level set of f through p.

Proof. We have 〈grad f |p, v〉 = vf , so the Cauchy-Schwarz inequality shows that

| 〈grad f |p, v〉g |

is maximized when v = ± grad f |p. If S is the level set of f through p then TpS = ker dfpby Proposition 5.38. Clearly 〈grad f |p, v〉 = vf = dfp(v) = 0 for every v ∈ ker dfp.

Theorem 234. [Problem 13-22] For any smooth manifold M with or without boundary,the vector bundles TM and T ∗M are smoothly isomorphic over M .

Proof. Choose any Riemannian metric g on M and let g : TM → T ∗M be the smoothbundle homomorphism defined at each p ∈M by g(v)(w) = gp(v, w). Then g is injective(and therefore bijective) at each point, so Theorem 174 shows that g is a smooth bundleisomorphism.

95

Example 235. [Problem 13-23] Is there a smooth covector field on S2 that vanishes atexactly one point?

Yes, due to Theorem 128 and Theorem 234.

Chapter 14. Differential Forms

Theorem 236. [Exercise 14.4] Let α be a covariant tensor on a finite-dimensionalvector space.

(1) Altα is alternating.(2) Altα = α if and only if α is alternating.

Proof. If τ ∈ Sk then

(Altα)(vτ(1), . . . , vτ(k)) =1

k!

∑σ∈Sk

(sgnσ)α(v(στ)(1), . . . , v(στ)(k))

= sgn(τ)1

k!

∑σ∈Sk

sgn(σ τ)α(v(στ)(1), . . . , v(στ)(k))

= sgn(τ)1

k!

∑σ∈Sk

(sgnσ)α(vσ(1), . . . , vσ(k))

= sgn(τ)(Altα)(v1, . . . , vk),

which proves (1). If α is alternating then

(Altα)(v1, . . . , vk) =1

k!

∑σ∈Sk

(sgnσ)α(vσ(1), . . . , vσ(k))

=1

k!

∑σ∈Sk

(sgnσ)2α(v1, . . . , vk)

= α(v1, . . . , vk).

Conversely, if Altα = α then α is alternating since Altα is alternating. This proves(2).

Theorem 237. [Exercise 14.12] The wedge product is the unique associative, bilinear,and anticommutative map Λk(V ∗)× Λl(V ∗)→ Λk+l(V ∗) satisfying

εi1 ∧ · · · ∧ εik = εI

for any basis (εi) for V ∗ and any multi-index I = (i1, . . . , ik).

96

Proof. Let ∧ be such a map. Let ω ∈ Λk(V ∗) and η ∈ Λl(V ∗). By Lemma 14.10,

ω∧η =

(′∑I

ωIεI

)∧

(′∑J

ηJεJ

)

=′∑I

′∑J

ωIηJεI∧εJ

=′∑I

′∑J

ωIηJεi1∧ · · · ∧eik∧ej1∧ · · · ∧ejl

=′∑I

′∑J

ωIηJεIJ

=′∑I

′∑J

ωIηJεI ∧ εJ

=

(′∑I

ωIεI

)∧

(′∑J

ηJεJ

)= ω ∧ η.

Theorem 238. [Exercise 14.14] ΛkT ∗M is a smooth subbundle of T kT ∗M , and there-fore is a smooth vector bundle of rank

(nk

)over M .

Proof. This follows from Proposition 14.8 and Lemma 10.32.

Theorem 239. [Exercise 14.28] The diagram below commutes:

C∞(R3) X(R3) X(R3) C∞(R3)

Ω0(R3) Ω1(R3) Ω2(R3) Ω3(R3).

grad curl div

d d d

Id [ β ∗

Therefore curl grad = 0 and div curl = 0 on R3. The analogues of the left-hand andright-hand square commute when R3 is replaced by Rn for any n.

97

Proof. In Rn, we have

([ grad)(f) = [

(n∑i=1

∂f

∂xi∂

∂xi

)=

n∑i=1

∂f

∂xidxi = df

and

(d β)(X) = d(Xy(dx1 ∧ · · · ∧ dxn))

= d

(n∑i=1

(−1)i−1dxi(X)dx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxn)

= d

(n∑i=1

(−1)i−1X idx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxn)

=n∑i=1

(−1)i−1

n∑j=1

∂X i

∂xjdxj ∧ dx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxn

=n∑i=1

∂X i

∂xidx1 ∧ · · · ∧ dxn

= ∗

(n∑i=1

∂X i

∂xi

)= (∗ div)(X).

In R3 we have

(β curl)(X) = β

((∂X3

∂x2− ∂X2

∂x3

)∂

∂x1+

(∂X1

∂x3− ∂X3

∂x1

)∂

∂x2

+

(∂X2

∂x1− ∂X1

∂x2

)∂

∂x3

)=

(∂X2

∂x1− ∂X1

∂x2

)dx1 ∧ dx2 +

(∂X3

∂x1− ∂X1

∂x3

)dx1 ∧ dx3

+

(∂X3

∂x2− ∂X2

∂x3

)dx2 ∧ dx3

= d(X1dx1 +X2dx2 +X3dx3

)= (d [)(X).

Theorem 240. [Problem 14-1] Covectors ω1, . . . , ωk ∈ V ∗ are linearly dependent if andonly if ω1 ∧ · · · ∧ ωk = 0.

98

Proof. Let n = dimV . Suppose that ω1, . . . , ωk are linearly dependent and assumewithout loss of generality that a1ω

1 + · · · + akωk = 0 with a1 6= 0. Then ω1 =

−a−11 (a2ω

2 + · · ·+ akωk), so

ω1 ∧ · · · ∧ ωk = −a−11 (a2ω

2 + · · ·+ akωk) ∧ ω2 ∧ · · · ∧ ωk = 0.

Conversely, suppose that ω1, . . . , ωk are linearly independent. We can extendω1, . . . , ωk

to a basis ω1, . . . , ωn of Λn(V ∗). Then

ω1 ∧ · · · ∧ ωn = (ω1 ∧ · · · ∧ ωk) ∧ (ωk+1 ∧ · · · ∧ ωn) 6= 0,

so ω1 ∧ · · · ∧ ωk 6= 0.

Theorem 241. [Problem 14-5] Let M be a smooth n-manifold with or without bound-ary, and let (ω1, . . . , ωk) be an ordered k-tuple of smooth 1-forms on an open subsetU ⊆M such that (ω1|p, . . . , ωk|p) is linearly independent for each p ∈ U . Given smooth1-forms α1, . . . , αk on U such that

k∑i=1

αi ∧ ωi = 0,

each αi can be written as a linear combination of ω1, . . . , ωk with smooth coefficients.

Proof. First note that each αi is in the span of ω1, . . . , ωk, since

αi ∧ ω1 ∧ · · · ∧ ωk = (−1)i+1ω1 ∧ · · · ∧ αi ∧ ωi ∧ · · · ∧ ωk

= (−1)iω1 ∧ · · · ∧∑j 6=i

αj ∧ ωj ∧ · · · ∧ ωk

= 0.

Therefore we have αi = rijωj for some coefficients rij : U → R. But the span of

ω1, . . . , ωk forms a smooth subbundle of TM , so Proposition 10.22 shows that thefunctions rij are smooth.

Example 242. [Problem 14-6] Define a 2-form ω on R3 by

ω = x dy ∧ dz + y dz ∧ dx+ z dx ∧ dy.

(1) Compute ω in spherical coordinates (ρ, ϕ, θ) defined by

(x, y, z) = (ρ sinϕ cos θ, ρ sinϕ sin θ, ρ cosϕ).

(2) Compute dω in both Cartesian and spherical coordinates and verify that bothexpressions represent the same 3-form.

(3) Compute the pullback ι∗S2ω to S2, using coordinates (ϕ, θ) on the open subsetwhere these coordinates are defined.

(4) Show that ι∗S2ω is nowhere zero.

99

We have

dx = sinϕ cos θ dρ+ ρ cosϕ cos θ dϕ− ρ sinϕ sin θ dθ,

dy = sinϕ sin θ dρ+ ρ cosϕ sin θ dϕ+ ρ sinϕ cos θ dθ,

dz = cosϕdρ− ρ sinϕdϕ,

so

ω = ρ3 sinϕdϕ ∧ dθ.Also,

dω = 3dx ∧ dy ∧ dzand

dω = 3ρ2 sinϕdρ ∧ dϕ ∧ dθ.The pullback of ω to S2 is given by

ι∗S2ω = sinϕdϕ ∧ dθ,

which is defined for (ϕ, θ) ∈ (0, π)×(0, 2π). This expression is never zero since sinϕ > 0for ϕ ∈ (0, π).

Chapter 15. Orientations

Theorem 243. [Exercise 15.4] Suppose M is an oriented smooth n-manifold with orwithout boundary, and n ≥ 1. Every local frame with connected domain is either posi-tively oriented or negatively oriented.

Proof. Let (Ei) be a local frame with a connected domain U . Let D be the points of

U at which (Ei) is positively oriented. Let p ∈ D and let (Ei) be an oriented local

frame defined on a neighborhood V ⊆ U of p. Writing Ei = Bji Ej for continuous

component functions Bji : V → R, we have det(Bj

i (p)) > 0. Since det is continuous,there is a neighborhood of p on which Ei is oriented. This shows that D is open. Asimilar argument shows that U \D is open, so D is either empty or equal to U by theconnectedness of U .

Theorem 244. [Exercise 15.8] Suppose M1, . . . ,Mk are orientable smooth manifolds.There is a unique orientation on M1 × · · · × Mk, called the product orientation,with the following property: if for each i = 1, . . . , k, ωi is an orientation form for thegiven orientation on Mi, then π∗1ω1 ∧ · · · ∧ π∗kωk is an orientation form for the productorientation.

100

Proof. Let η1, . . . , ηk be orientation forms for M1, . . . ,Mk and let O be the orientationdetermined by π∗1η1∧· · ·∧π∗kηk. Suppose ω1, . . . , ωk are orientation forms forM1, . . . ,Mk.Then ωi = fiηi for some strictly positive continuous functions fi : Mi → R, so

π∗1ω1 ∧ · · · ∧ π∗kωk = (f1 π1) · · · (fk πk)η1 ∧ · · · ∧ ηk.

Since (f1 π1) · · · (fk πk) is strictly positive, π∗1ω1 ∧ · · · ∧ π∗kωk determines the sameorientation as η1 ∧ · · · ∧ ηk.

Theorem 245. [Exercise 15.10] Let M be a connected, orientable, smooth manifoldwith or without boundary. Then M has exactly two orientations. If two orientations ofM agree at one point, they are equal.

Proof. If dimM = 0 then M consists of a single point, so the result is clear. SupposeO,O′ are two orientations ofM that agree at a single point p. LetD = x ∈M : Ox = O′x;this set is nonempty since p ∈ D. Let x ∈ D and let (Ei) be a local frame defined ina connected neighborhood of x that is positively oriented with respect to O′. Apply-ing Theorem 243 to (M,O) shows that (Ei) is positively oriented with respect to O.Therefore D is open. A similar argument shows that M \D is open, so D = M by theconnectedness of M .

Since M is orientable, there is an orientation O for M . Let ω be an orientation formfor O; then −ω determines an orientation O′ for M with Op 6= O′p for all p ∈ M . IfO′′ is any orientation and p is any point in M then either O′′p = Op or O′′p = O′p, soO′′ = O or O′′ = O′.

Theorem 246. [Exercise 15.12] Suppose M is an oriented smooth manifold with orwithout boundary, and D ⊆M is a smooth codimension-0 submanifold with or withoutboundary. The orientation of M restricts to an orientation of D. If ω is an orientationform for M , then ι∗Dω is an orientation form for D.

Proof. It suffices to check that ι∗Dω is nonvanishing. But this is clear since d(ιD)p is anisomorphism for every p ∈M .

Theorem 247. [Exercise 15.13] Suppose M and N are oriented positive-dimensionalsmooth manifolds with or without boundary, and F : M → N is a local diffeomorphism.The following are equivalent:

(1) F is orientation-preserving.(2) With respect to any oriented smooth charts for M and N , the Jacobian matrix

of F has positive determinant.(3) For any positively oriented orientation form ω for N , the form F ∗ω is positively

oriented for M .

101

Proof. If (U, (xi)) and (V, (yi)) are oriented smooth charts for M and N respectively,then dFp takes the oriented basis (∂/∂xi|p) to an oriented basis(

∂F j

∂xi(p)

∂

∂yj

∣∣∣∣p

).

Since (∂/∂yi|p) is also an oriented basis, we have det(∂F j/∂xi(p)) > 0. This proves

(1) ⇒ (2). Suppose that (2) holds. By shrinking U and V , we can assume that theyare connected sets and that F |U : U → V is a diffeomorphism. Then

ω = f dy1 ∧ · · · ∧ dyn

on V for some positive continuous function f , so by Proposition 14.20 we have

F ∗ω = (f F )(detDF )dx1 ∧ · · · ∧ dxn.Since detDF > 0 by (2), F ∗ω is positively oriented on U . This proves (2) ⇒ (3).Finally, suppose that (3) holds and let (Ei) be an oriented basis of TpM . If ω is apositively oriented orientation form for N then

(F ∗ω)p(E1, . . . , En) > 0⇒ ωF (p)(dFp(E1), . . . , dFp(En)) > 0,

so (dFp(Ei)) is an oriented basis of TF (p)N . This proves (3)⇒ (1).

Theorem 248. [Exercise 15.14] A composition of orientation-preserving maps is orientation-preserving.

Proof. Let F : M → N and G : N → P be orientation-preserving maps. If (Ei) is anoriented basis of TpM then (dFp(Ei)) is an oriented basis of TF (p)N and (d(GF )p(Ei))is an oriented basis of T(GF )(p)P .

Theorem 249. [Exercise 15.16] Suppose F : M → N and G : N → P are localdiffeomorphisms and O is an orientation on P . Then (G F )∗O = F ∗(G∗O).

Proof. This is clear from the fact that (G F )∗ω = F ∗G∗ω for any orientation formω.

Theorem 250. [Exercise 15.20] Every Lie group has precisely two left-invariant ori-entations, corresponding to the two orientations of its Lie algebra.

Proof. Let G be a Lie group. Corollary 8.39 shows that there are at least two left-invariant orientations. If O is a left-invariant orientation on G then it is completelydetermined by Oe, so there are exactly two left-invariant orientations.

Theorem 251. [Exercise 15.30] Suppose (M, g) and (M, g) are positive-dimensional

Riemannian manifolds with or without boundary, and F : M → M is a local isometry.Then F ∗ωg = ωg.

102

Proof. Let p ∈ M and let U be a neighborhood of p for which F |U is an isometry.If (Ei) is a local oriented orthonormal frame for M then (dF (Ei)) is a local oriented

orthonormal frame for M on a neighborhood of F (p) since

〈dF (Ei), dF (Ej)〉g = 〈Ei, Ej〉g = δij.

Therefore(F ∗ωg)(E1, . . . , En) = ωg(dF (E1), . . . , dF (En)) = 1,

which shows that F ∗ωg is the Riemannian volume form on M .

Theorem 252. [Problem 15-1] Suppose M is a smooth manifold that is the union oftwo orientable open submanifolds with connected intersection. Then M is orientable.

Proof. Denote the two open submanifolds by N and P . We can assume that there issome p ∈ N ∩P , for otherwise the result is trivial. Choose an orientation O for N , andchoose an orientation O′ for P such that Op = O′p. By Theorem 245, O and O′ agreeon N ∩ P , and it is easy to check that the orientation given by

O′′x =

Ox, x ∈ N,O′x, x ∈ P

is well-defined and continuous.

Theorem 253. [Problem 15-2] Suppose M and N are oriented smooth manifolds withor without boundary, and F : M → N is a local diffeomorphism. If M is connected,then F is either orientation-preserving or orientation-reversing.

Proof. If M and N are 0-manifolds, then M and N both consist of a single point andthe result is trivial. Let D be the set of points x for which dFx takes oriented bases ofTxM to oriented bases of TF (x)N . Let ω be a positively oriented orientation form forN . Let x ∈ D and let (Ei) be an oriented local frame defined on a neighborhood U ofx. Then (F ∗ω)x(E1|x, . . . , En|x) > 0, so by continuity there is a neighborhood V ⊆ Uof x on which (F ∗ω)(E1, . . . , En) > 0. Therefore V ⊆ D by Theorem 247, and D isopen. A similar argument shows that M \D is open, so D = ∅ or D = M since M isconnected.

Theorem 254. [Problem 15-3] Suppose n ≥ 1, and let α : Sn → Sn be the antipodalmap: α(x) = −x. Then α is orientation-preserving if and only if n is odd.

Proof. The outward unit normal vector field along Sn is N = xi∂/∂xi, and Corollary15.34 shows that the Riemannian volume form ωg of Sn is

ωg = ι∗Sn(Ny(dx1 ∧ · · · ∧ dxn+1)) = ι∗Sn

(n+1∑i=1

(−1)i−1xi dx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxn+1

).

103

We have

α∗ωg = ι∗Sn

(n+1∑i=1

(−1)i−1(−xi) (−dx1) ∧ · · · ∧ (−dxi) ∧ · · · ∧ (−dxn+1)

)= (−1)n+1ωg,

so F is orientation-preserving if and only if n is odd.

Theorem 255. [Problem 15-5] Let M be a smooth manifold with or without boundary.The total spaces of TM and T ∗M are orientable.

Proof. Let π : TM → M be the projection. Let (U,ϕ) and (V, ψ) be smooth charts

for M , and let (π−1(U), ϕ) and (π−1(V ), ψ) be the corresponding charts for TM . Thetransition map is given by

(ψ ϕ−1)(x1, . . . , xn, v1, . . . , vn) =

(x1(x), . . . , xn(x),

∂x1

∂xj(x)vj, . . . ,

∂xn

∂xj(x)vj

).

The Jacobian matrix of this map at (x1, . . . , xn, v1, . . . , vn) is[∂xi

∂xj(x) 0

∗ ∂xi

∂xj(x)

],

which clearly has positive determinant. The result follows from Proposition 15.6. Asimilar computation holds for T ∗M .

Theorem 256. [Problem 15-7] Suppose M is an oriented Riemannian manifold withor without boundary, and S ⊆ M is an oriented smooth hypersurface with or withoutboundary. There is a unique smooth unit normal vector field along S that determinesthe given orientation of S.

Proof. Let n = dimM , let ω be an orientation form for M , and let O be the orientationon S. Let NS be the normal bundle of S (see Theorem 213), which is of rank 1. Let Zbe the set of all zero vectors in NS. Define a map f : NS \ Z → −1,+1 as follows:if v ∈ NpS then choose an oriented (with respect to O) basis (Ei) for TpS and define

f(v) =ω(v, E1, . . . , En−1)

|ω(v, E1, . . . , En−1)|.

This map is well-defined because (Ei) is oriented, and is continuous because there isan oriented local frame defined on a neighborhood of every p. Let P = f−1(+1). Itis easy to see that P is convex and open in NS, so Theorem 217 shows that there is

a smooth global section X : S → P . Then X = X/ |X|g is the desired smooth unitnormal vector field along S.

104

Theorem 257. [Problem 15-8] Suppose M is an orientable Riemannian manifold, andS ⊆M is an immersed or embedded submanifold with or without boundary.

(1) If S has trivial normal bundle, then S is orientable.(2) If S is an orientable hypersurface, then S has trivial normal bundle.

Proof. If S has trivial normal bundle then there is a global frame for NS, and Proposi-tion 15.21 shows that S is orientable. If S is an orientable hypersurface, then Theorem256 gives a global frame for NS (since NS is of rank 1).

Theorem 258. [Problem 15-9] Let S be an oriented, embedded, 2-dimensional subman-ifold with boundary in R3, and let C = ∂S with the induced orientation. By Theorem256, there is a unique smooth unit normal vector field N on S that determines the ori-entation. Let T be the oriented unit tangent vector field on C, and let V be the uniqueunit vector field tangent to S along C that is orthogonal to T and inward-pointing. Then(Tp, Vp, Np) is an oriented orthonormal basis for R3 at each p ∈ C.

Proof. It is clear that (Tp, Vp, Np) is an orthonormal basis, so it remains to check that(Tp, Vp, Np) is oriented. This is true if and only if (Np, Tp, Vp) is oriented, and by thedefinition of the induced orientation, this is true if and only if (Tp, Vp) is an orientedbasis for TpS. This is true since (−Vp, Tp) is an oriented basis for TpS.

Theorem 259. [Problem 15-10] Let M be a connected nonorientable smooth manifold

with or without boundary, and let π : M → M be its orientation covering. If X isany oriented smooth manifold with or without boundary, and F : X → M is any localdiffeomorphism, then there exists a unique orientation-preserving local diffeomorphism

F : X → M such that π F = F .

Proof. Define F (x) = (F (x),OF (x)), where OF (x) is the orientation of TF (x)M deter-

mined by F and the orientation of X. It is clear that π F = F . Let UO be an element

of the basis defined in Proposition 15.40. If U is connected, then F−1(UO) is either U or

empty. Therefore F−1(UO) is open for any open set U , and F is continuous. By Theo-

rem 41, F is smooth, and it is easy to check that it is orientation-preserving and a local

diffeomorphism. Uniqueness follows from the fact that F is orientation-preserving.

Theorem 260. [Problem 15-11] Let M be a nonorientable connected smooth manifold

with or without boundary, and let π : M → M be its orientation covering. If Mis an oriented smooth manifold with or without boundary that admits a two-sheeted

smooth covering map π : M → M , then there exists a unique orientation-preserving

diffeomorphism ϕ : M → M such that π ϕ = π.

105

Proof. By Theorem 259, there is a unique orientation-preserving local diffeomorphism

ϕ : M → M such that π ϕ = π. Similarly, there is a unique orientation-preserving

local diffeomorphism ψ : M → M such that π ψ = π. Then π ϕ ψ = π ψ = π, soϕψ = IdM by uniqueness. Similarly, ψϕ = IdM . Therefore ϕ is a diffeomorphism.

Theorem 261. [Problem 15-22] Every orientation-reversing diffeomorphism of R hasa fixed point.

Proof. Let f : R → R be such a diffeomorphism; then f ′(x) < 0 for all x ∈ R. Letg(x) = f(x) − x; then g′(x) = f ′(x) − 1 < −1 for all x ∈ R. If g(0) = 0, then we aredone. If g(0) > 0 then by the mean value theorem, there exists a point t ∈ (0, g(0))such that

g(g(0))− g(0)

g(0)= g′(t) < −1,

i.e. g(g(0)) < 0, and the intermediate value theorem shows that g(x) = 0 for somex ∈ (0, g(0)). Similarly, if g(0) < 0 then there exists a point t ∈ (g(0), 0) such that

g(g(0))− g(0)

g(0)= g′(t) < −1,

i.e. g(g(0)) > 0, and the intermediate value theorem shows that g(x) = 0 for somex ∈ (g(0), 0).

Chapter 16. Integration on Manifolds

Theorem 262. [Exercise 16.29] Suppose (M, g) is an oriented Riemannian manifoldand f : M → R is continuous and compactly supported. Then∣∣∣∣ˆ

M

f dVg

∣∣∣∣ ≤ ˆM

|f | dVg.

Proof. If f is supported in the domain of a single oriented smooth chart (U,ϕ), then∣∣∣∣ˆM

f dVg

∣∣∣∣ =

∣∣∣∣ˆϕ(U)

f(x)√

det(gij) dx1 · · · dxn

∣∣∣∣≤ˆϕ(U)

|f(x)|√

det(gij) dx1 · · · dxn

=

ˆM

|f | dVg.

106

For the general case, cover supp f by finitely many domains of positively or negativelyoriented smooth charts Ui and choose a subordinate smooth partition of unity ψi.Then ∣∣∣∣ˆ

M

f dVg

∣∣∣∣ =

∣∣∣∣∣∑i

ˆM

ψif dVg

∣∣∣∣∣≤∑i

ˆM

ψi |f | dVg

=

ˆM

|f | dVg.

Example 263. [Problem 16-2] Let T2 = S1×S1 ⊆ R4 denote the 2-torus, defined as theset of points (w, x, y, z) such that w2 + x2 = y2 + z2 = 1, with the product orientationdetermined by the standard orientation on S1. Compute

´T2 ω, where ω is the following

2-form on R4:

ω = xyz dw ∧ dy.

Let D = (0, 2π)× (0, 2π) and define an orientation-preserving diffeomorphism F : D →T2 by

F (θ, ϕ) = (cos θ, sin θ, cosϕ, sinϕ)

so that

dw = − sin θ dθ, dx = cos θ dθ,

dy = − sinϕdϕ, dz = cosϕdϕ.

Then ˆT2

ω =

ˆT2

sin θ cosϕ sinϕ(− sin θ dθ) ∧ (− sinϕdϕ)

=

ˆT2

sin2 θ cosϕ sin2 ϕdθ ∧ dϕ

=

(ˆ 2π

0

sin2 θ dθ

)(ˆ 2π

0

cosϕ sin2 ϕdϕ

)= π · 0= 0.

Theorem 264. [Problem 16-3] Suppose E and M are smooth n-manifolds with orwithout boundary, and π : E → M is a smooth k-sheeted covering map or generalizedcovering map.

107

(1) If E and M are oriented and π is orientation-preserving, then´Eπ∗ω = k

´Mω

for any compactly supported n-form ω on M .(2)´Eπ∗µ = k

´Mµ whenever µ is a compactly supported density on M .

Proof. Let Ui be a finite open cover of suppω by evenly covered sets, and let ψi be a

subordinate smooth partition of unity. For each i, let U(1)i , . . . , U

(k)i be the components

of π−1(Ui). Letψ

(j)i

be a smooth partition of unity subordinate to the open cover

U(j)i

of supp π∗ω. Applying Proposition 16.6(d), we have

k

ˆM

ω =∑i

k∑j=1

ˆUi

ψiω

=∑i

k∑j=1

ˆU

(j)i

(ψi π|U(j)i

)π∗ω

=∑i

k∑j=1

ˆU

(j)i

(ψi π|U(j)i

)π∗ω

=∑i

k∑j=1

ˆU

(j)i

∑i′

k∑j′=1

ψ(j′)i′ (ψi π|U(j)

i)π∗ω

=

ˆE

∑i′

k∑j′=1

ψ(j′)i′

∑i

k∑j=1

(ψi π|U(j)i

)π∗ω

=

ˆE

∑i′

k∑j′=1

ψ(j′)i′ π

∗ω

=

ˆE

π∗ω.

This proves (1). Part (2) is similar.

Theorem 265. [Problem 16-4] If M is an oriented compact smooth manifold withboundary, then there does not exist a retraction of M onto its boundary.

Proof. Suppose that there exists such a retraction r : M → ∂M . By Theorem 6.26, wecan assume that r is smooth. Let ι : ∂M → M be the inclusion map. Let ω be anorientation form for ∂M . By Theorem 16.11 and Proposition 14.26,ˆ

∂M

r∗ω =

ˆM

d(r∗ω) =

ˆM

r∗(dω) = 0

108

since ω is a top-degree form. By definition,ˆ∂M

r∗ω =

ˆ∂M

ι∗r∗ω =

ˆ∂M

(r ι)∗ω =

ˆ∂M

ω,

so ˆ∂M

ω = 0.

This contradicts Proposition 16.6(c).

Theorem 266. [Problem 16-5] Suppose M and N are oriented, compact, connected,smooth manifolds, and F,G : M → N are homotopic diffeomorphisms. Then F and Gare either both orientation-preserving or both orientation-reversing.

Proof. Theorem 253 shows that F and G are individually orientation-preserving ororientation-reversing. By Theorem 6.29, there is a smooth homotopy H : M × I → Nfrom F to G. Let ω be a positively oriented orientation form for N . By Theorem 16.11,ˆ

∂(M×I)H∗ω =

ˆM×I

d(H∗ω) =

ˆM×I

H∗(dω) = 0.

But ˆ∂(M×I)

H∗ω =

ˆM

F ∗ω −ˆM

G∗ω,

so ˆM

F ∗ω =

ˆM

G∗ω.

The result follows from Proposition 16.6(d).

Theorem 267. [Problem 16-6] The following are equivalent:

(1) There exists a nowhere-vanishing vector field on Sn.(2) There exists a continuous map V : Sn → Sn satisfying V (x) ⊥ x (with respect

to the Euclidean dot product on Rn+1) for all x ∈ Sn.(3) The antipodal map α : Sn → Sn is homotopic to IdSn.(4) The antipodal map α : Sn → Sn is orientation-preserving.(5) n is odd.

Therefore, there exists a nowhere-vanishing vector field on Sn if and only if n is odd.

Proof. Suppose X is a nowhere-vanishing vector field on Sn. Then taking V = X/ |X|proves (1)⇒ (2). Suppose (2) holds. We can define a homotopy H1 : Sn× I → Sn fromα to V by

H1(x, t) =(1− t)α(x) + tV (x)

|(1− t)α(x) + tV (x)|,

109

where the denominator is nonzero since (1 − t)(−x) + tV (x) = 0 contradicts the factthat V (x) ⊥ x. Similarly, we can define a homotopy H2 from V to IdSn by

H2(x, t) =(1− t)V (x) + tx

|(1− t)V (x) + tx|.

This proves (2) ⇒ (3). Theorem 266 proves (3) ⇒ (4), and Theorem 254 proves(4)⇒ (5). Finally, Theorem 150 proves (5)⇒ (1).

Theorem 268. [Problem 16-9] Let ω be the (n− 1)-form on Rn \ 0 defined by

ω = |x|−nn∑i=1

(−1)i−1xi dx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxn.

(1) ι∗Sn−1ω is the Riemannian volume form of Sn−1 with respect to the round metricand the standard orientation.

(2) ω is closed but not exact on Rn \ 0.

Proof. The unit normal vector field for Sn−1 with the standard orientation is given byx 7→ x, so (1) follows from the formula in Lemma 14.13. For (2), we have

dω =n∑i=1

(−1)i−1d

(xi

|x|n)dx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxn

=n∑i=1

(|x|n − n(xi)2 |x|n−2

|x|2n

)dx1 ∧ · · · ∧ dxn

=n∑i=1

(1

|x|n− n(xi)2

|x|n+2

)dx1 ∧ · · · ∧ dxn

= 0.

If ω is exact then ˆSn−1

ω = 0

by Corollary 16.13, which is a contradiction.

Example 269. [Problem 16-10] Let D denote the torus of revolution in R3 obtainedby revolving the circle (r−2)2 +z2 = 1 around the z-axis, with its induced Riemannianmetric and with the orientation determined by the outward unit normal.

(1) Compute the surface area of D.(2) Compute the integral over D of the function f(x, y, z) = z2 + 1.(3) Compute the integral over D of the 2-form ω = z dx ∧ dy.

110

Let ι : D → R3 be the inclusion map. Let U = (0, 2π) × (0, 2π) and define anorientation-preserving diffeomorphism F : U → D by

F (θ, ϕ) = ((2 + cos θ) cosϕ, (2 + cos θ) sinϕ, sin θ).

We have

dx = − sin θ cosϕdθ − (2 + cos θ) sinϕdϕ,

dy = − sin θ sinϕdθ + (2 + cos θ) cosϕdϕ,

dz = cos θ dθ

and

F ∗g = dx2 + dy2 + dz2

= sin2 θ cos2 ϕdθ2 + (2 + cos θ)2 sin2 ϕdϕ2

+ 2 sin θ sinϕ cosϕ(2 + cos θ) dθ dϕ

+ sin2 θ sin2 ϕdθ2 + (2 + cos θ)2 cos2 ϕdϕ2

− 2 sin θ sinϕ cosϕ(2 + cos θ) dθ dϕ

+ cos2 θ dθ2

= dθ2 + (2 + cos θ)2 dϕ2.

For (1), we have ˆD

ωι∗g =

ˆ(θ,ϕ)∈U

√(2 + cos θ)2

= 2π

ˆ 2π

0

(2 + cos θ) dθ

= 8π2.

For (2), we have ˆD

fωι∗g =

ˆ(θ,ϕ)∈U

(sin2 θ + 1)√

(2 + cos θ)2

= 2π

ˆ 2π

0

(sin2 θ + 1)(2 + cos θ) dθ

= 12π2.

For (3), we have

dx ∧ dy = (− sin θ cosϕdθ − (2 + cos θ) sinϕdϕ)

∧ (− sin θ sinϕdθ + (2 + cos θ) cosϕdϕ)

= −(2 + cos θ) sin θ dθ ∧ dϕ

111

so that ˆD

z dx ∧ dy = −ˆD

(2 + cos θ) sin2 θ dθ ∧ dϕ

= −2π

ˆ 2π

0

(2 + cos θ) sin2 θ dθ

= −4π2.

Theorem 270. [Problem 16-11] Let (M, g) be a Riemannian n-manifold with or withoutboundary. In any smooth local coordinates (xi) we have

div

(X i ∂

∂xi

)=

1√det g

∂

∂xi

(X i√

det g),

where det g = det(gkl) is the determinant of the component matrix of g in these coordi-nates.

Proof. We have

ωg =√

det gdx1 ∧ · · · ∧ dxn

by Proposition 15.31, so

β

(X i ∂

∂xi

)=

n∑i=1

(−1)i−1X i√

det g dx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxn.

Then

d

(β

(X i ∂

∂xi

))=

∂

∂xi

(X i√

det g)dx1 ∧ · · · ∧ dxn

and

div

(X i ∂

∂xi

)= ∗−1

(d

(β

(X i ∂

∂xi

)))=

1√det g

∂

∂xi

(X i√

det g).

Theorem 271. [Problem 16-12] Let (M, g) be a compact Riemannian manifold withboundary, let g denote the induced Riemannian metric on ∂M , and let N be the outwardunit normal vector field along ∂M .

(1) The divergence operator satisfies the following product rule for f ∈ C∞(M),X ∈ X(M):

div(fX) = f divX + 〈grad f,X〉g .

112

(2) We have the following “integration by parts” formula:ˆM

〈grad f,X〉g dVg =

ˆ∂M

f 〈X,N〉g dVg −ˆM

(f divX) dVg.

Proof. Theorem 270 shows that in coordinates, we have

div(fX) =1√

det g

∂

∂xi

(fX i

√det g

)=

1√det g

(f∂

∂xi

(X i√

det g)

+∂f

∂xiX i√

det g

)= f

1√det g

∂

∂xi

(X i√

det g)

+∂f

∂xiX i

= f divX + 〈grad f,X〉g .

Therefore ˆM

〈grad f,X〉g dVg =

ˆM

div(fX) dVg −ˆM

f divX dVg

=

ˆ∂M

f 〈X,N〉g dVg −ˆM

f divX dVg

by Theorem 16.32.

Theorem 272. [Problem 16-13] Let (M, g) be a Riemannian n-manifold with or withoutboundary. The linear operator 4 : C∞(M) → C∞(M) defined by 4u = − div(gradu)is called the (geometric) Laplacian. The Laplacian is given in any smooth localcoordinates by

4u = − 1√det g

∂

∂xi

(gij√

det g∂u

∂xj

).

On Rn with the Euclidean metric and standard coordinates,

4u = −n∑i=1

∂2u

(∂xi)2.

Proof. Obvious from Theorem 270.

Theorem 273. [Problem 16-14] Let (M, g) be a Riemannian n-manifold with or withoutboundary. A function u ∈ C∞(M) is said to be harmonic if 4u = 0.

(1) If M is compact, thenˆM

u4v dVg =

ˆM

〈gradu, grad v〉g dVg −ˆ∂M

uNv dVg,

113

ˆM

(u4v − v4u) dVg =

ˆ∂M

(vNu− uNv)dVg,

where N and g are as in Theorem 271. These are known as Green’s identities.(2) If M is compact and connected and ∂M = ∅, the only harmonic functions on

M are the constants.(3) If M is compact and connected, ∂M 6= ∅, and u, v are harmonic functions on

M whose restrictions to ∂M agree, then u = v.

Proof. (1) follows by putting f = u and X = grad v in Theorem 271. For (2), if u isharmonic then ˆ

M

|gradu|2g dVg =

ˆM

u4u dVg = 0

by part (1). Therefore gradu = 0, and u is constant since M is connected. For (3), wehave ˆ

M

|grad(u− v)|2g dVg =

ˆ∂M

(u− v)N(u− v) dVg = 0

by part (1). Therefore grad(u − v) = 0, and u − v is constant since M is connected.Since u and v agree on ∂M , we have u = v.

Theorem 274. [Problem 16-15] Let (M, g) be a compact connected Riemannian man-ifold without boundary, and let 4 be its geometric Laplacian. A real number λ is calledan eigenvalue of 4 if there exists a smooth real-valued function u on M , not identicallyzero, such that 4u = λu. In this case, u is called an eigenfunction corresponding toλ.

(1) 0 is an eigenvalue of 4, and all other eigenvalues are strictly positive.(2) If u and v are eigenfunctions corresponding to distinct eigenvalues, then

´Muv dVg =

0.

Proof. Any constant function is an eigenfunction with eigenvalue 0. If 4u = λu and uis not constant then Theorem 273 shows that

0 <

ˆM

|gradu|2g dVg =

ˆM

u4u dVg = λ

ˆM

u2 dVg,

so λ > 0. Suppose 4u = λu and 4v = µv where λ 6= µ. By Theorem 273,

0 =

ˆM

(u4v − v4u) dVg = (µ− λ)

ˆM

uv dVg,

so ˆM

uv dVg = 0

since µ− λ 6= 0.

114

Theorem 275. [Problem 16-16] Let M be a compact connected Riemannian n-manifoldwith nonempty boundary. A number λ ∈ R is called a Dirichlet eigenvalue for Mif there exists a smooth real-valued function u on M , not identically zero, such that4u = λu and u|∂M = 0. Similarly, λ is called a Neumann eigenvalue if there existssuch a u satisfying 4u = λu and Nu|∂M = 0, where N is the outward unit normal.

(1) Every Dirichlet eigenvalue is strictly positive.(2) 0 is a Neumann eigenvalue, and all other Neumann eigenvalues are strictly

positive.

Proof. As in Theorem 274.

Theorem 276. [Problem 16-17] Suppose M is a compact connected Riemannian n-manifold with nonempty boundary. A function u ∈ C∞(M) is harmonic if and onlyif it minimizes

´M|gradu|2g dVg among all smooth functions with the same boundary

values.

Proof. Suppose f ∈ C∞(M) vanishes on ∂M . For all ε, we haveˆM

|grad(u+ εf)|2g dVg =

ˆM

|gradu|2g dVg + 2ε

ˆM

〈gradu, grad f〉g dVg

+ ε2

ˆM

|grad f |2g dVg.

Applying Theorem 271, we haveˆM

|grad(u+ εf)|2g dVg =

ˆM

|gradu|2g dVg + 2ε

ˆM

(f4u) dVg

+ ε2

ˆM

|grad f |2g dVg.

If u is harmonic thenˆM

|grad(u+ f)|2g dVg =

ˆM

|gradu|2g dVg +

ˆM

|grad f |2g dVg,

and it is clear that u minimizes´M|gradu|2g dVg among all smooth functions with the

same boundary values. Conversely, suppose that u minimizes´M|gradu|2g dVg. Thenˆ

M

|grad(u+ εf)|2g dVg

has a global minimum at ε = 0, and its derivative (with respect to ε)

2

ˆM

(f4u) dVg + 2ε

ˆM

|grad f |2g dVg

115

must be zero at ε = 0. That is, ˆM

(f4u) dVg = 0

for all f ∈ C∞(M) such that f |∂M = 0. Therefore 4u = 0.

Theorem 277. [Problem 16-18] Let (M, g) be an oriented Riemannian n-manifold.

(1) For each k = 1, . . . , n, g determines a unique inner product on Λk(T ∗pM) (de-noted by 〈·, ·〉g, just like the inner product on TpM) satisfying⟨

ω1 ∧ · · · ∧ ωk, η1 ∧ · · · ∧ ηk⟩g

= det(⟨

(ωi)#, (ηj)#⟩g

)whenever ω1, . . . , ωk, η1, . . . , ηk are covectors at p.

(2) The Riemannian volume form dVg is the unique positively oriented n-form thathas unit norm with respect to this inner product.

(3) For each k = 0, . . . , n, there is a unique smooth bundle homomorphism ∗ :ΛkT ∗M → Λn−kT ∗M satisfying

ω ∧ ∗η = 〈ω, η〉g dVg

for all smooth k-forms ω, η. (For k = 0, interpret the inner product as ordinarymultiplication.) This map is called the Hodge star operator.

(4) ∗ : Λ0T ∗M → ΛnT ∗M is given by ∗f = f dVg.(5) ∗ ∗ ω = (−1)k(n−k)ω if ω is a k-form.

Proof. For (1), any such inner product must satisfy

⟨εI , εJ

⟩g

= det

⟨(εi1)#, (εj1)#

⟩g· · ·

⟨(εi1)#, (εjk)#

⟩g

.... . .

...⟨(εik)#, (εj1)#

⟩g· · ·

⟨(εik)#, (εjk)#

⟩g

= det

gi1j1 · · · gi1jk...

. . ....

gikj1 · · · gikjk

= δIJ(*)

for all increasing multi-indices I and J , whenever (εi) is the coframe dual to a localorthonormal frame (Ei). (Here (gij) is the inverse of gij(p), which is the matrix of gin these coordinates.) This proves that the inner product is uniquely determined. Toprove existence, we will define 〈·, ·〉g by (*) and extend bilinearly. To check that the

116

definition is independent of the choice of orthonormal frame, let (εi) be another coframe

dual to a local orthonormal frame (Ei). Then

εI =′∑L

αLεL and εJ =

′∑M

βMεM

where αL = εI(El1 , . . . , Elk) and βM = εJ(Em1 , . . . , Emk), so⟨εI , εJ

⟩g

=′∑L

′∑M

αLβM⟨εL, εM

⟩g

=′∑L

′∑M

αLβMδLM

=′∑L

αLβL

=′∑L

εI(El1 , . . . , Elk)εJ(El1 , . . . , Elk).

Since Ei = Aji Ej for some orthonormal matrix (Aji ), we have⟨εI , εJ

⟩g

=′∑L

∑P

Ap1l1 · · ·ApklkδIP∑Q

Aq1l1 · · ·AqklkδJQ

=′∑L

∑σ∈Sk

(sgnσ)Aiσ(1)l1· · ·Aiσ(k)lk

∑τ∈Sk

(sgn τ)Ajτ(1)l1· · ·Ajτ(k)lk

=???????????????

For (2), if ω is any positively oriented n-form then ωp = c(dVg)p for some c > 0, and

〈ωp, ωp〉g = c2 〈dVg, dVg〉g= c2

⟨ε1 ∧ · · · ∧ εn, ε1 ∧ · · · ∧ εn

⟩g

= c2

whenever (εi) is the coframe dual to a local orthonormal frame, by Proposition 15.29.If ω has unit norm then it is clear that c = 1.

Example 278. [Problem 16-19] Consider Rn as a Riemannian manifold with the Eu-clidean metric and the standard orientation.

(1) Calculate ∗dxi for i = 1, . . . , n.(2) Calculate ∗(dxi ∧ dxj) in the case n = 4.

117

TODO

Theorem 279. [Problem 16-20] Let M be an oriented Riemannian 4-manifold. A2-form ω on M is said to be self-dual if ∗ω = ω, and anti-self-dual if ∗ω = −ω.

(1) Every 2-form ω on M can be written uniquely as a sum of a self-dual form andan anti-self-dual form.

(2) On M = R4 with the Euclidean metric, the self-dual and anti-self-dual forms instandard coordinates are given by:

??????????

Proof. TODO

Theorem 280. [Problem 16-21] Let (M, g) be an oriented Riemannian manifold andX ∈ X(M). Then

XydVg = ∗X[,

divX = ∗d ∗X[,

and, when dimM = 3,

curlX = (∗dX[)].

Proof. TODO

Theorem 281. [Problem 16-22] Let (M, g) be a compact, oriented Riemannian n-manifold. For 1 ≤ k ≤ n, define a map d∗ : Ωk(M)→ Ωk−1(M) by d∗ω = (−1)n(k+1)+1∗d∗ω, where ∗ is the Hodge star operator defined in Theorem 277. Extend this definitionto 0-forms by defining d∗ω = 0 for ω ∈ Ω0(M).

(1) d∗ d∗ = 0.(2) The formula

(ω, η) =

ˆM

〈ω, η〉g dVg

defines an inner product on Ωk(M) for each k, where 〈·, ·〉g is the pointwise innerproduct on forms defined in Theorem 277.

(3) (d∗ω, η) = (ω, dη) for all ω ∈ Ωk(M) and η ∈ Ωk−1(M).

Proof. TODO

118

Chapter 17. De Rham Cohomology

Theorem 282. [Exercise 17.37] Suppose M , N , and P are compact, connected, ori-ented, smooth n-manifolds.

(1) If F : M → N and G : N → P are both smooth maps, then deg(G F ) =(degG)(degF ).

(2) If F : M → N is a diffeomorphism, then degF = +1 if F is orientation-preserving and −1 if it is orientation-reversing.

(3) If two smooth maps F0, F1 : M → N are homotopic, then they have the samedegree.

Proof. For every smooth n-form ω on P , we haveˆM

(G F )∗ω =

ˆM

F ∗G∗ω

= (degF )

ˆN

G∗ω

= (degF )(degG)

ˆP

ω.

This proves (1). Part (2) follows from Proposition 16.6(d). If F0 and F1 are homotopicthen Proposition 17.10 shows that the induced cohomology maps F ∗0 and F ∗1 are equal.For any smooth n-form ω on N , we have F ∗0ω−F ∗1ω = dη for some smooth (n−1)-formη. Then ˆ

M

F ∗0ω −ˆM

F ∗1ω =

ˆM

dη = 0.

Theorem 283. [Problem 17-1] Let M be a smooth manifold with or without boundary,and let ω ∈ Ωp(M), η ∈ Ωq(M) be closed forms. The de Rham cohomology class ofω∧η depends only on the cohomology classes of ω and η, and thus there is a well-definedbilinear map `: Hp

dR(M) × HqdR(M) → Hp+q

dR (M), called the cup product, given by[ω] ` [η] = [ω ∧ η].

Proof. TODO

References

[1] John M. Lee. Introduction to Topological Manifolds. Springer, 2nd edition, 2011.