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, ) 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|s1 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 : U Bn centered at some p U .Let A be the smooth atlas obtained by replacing every coordinate map : V R inA with : V (M \ {p}) R, except when = . For any s > 0, let As be thesmooth atlas obtained from A by replacing with Fs . 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 : U R such that ei(z) = z for all z U . There exists an angle function on an open subset U S1 if and only if U 6= S1. For any such angle function, (U, ) 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 : S1 \ {N} R be thestereographic projection given by (x1, x2) = x1/(1 x2). We can compute

( 1)() = cos1 sin

,

which is a diffeomorphism on (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 (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 (S) represents the linear map (Q|S) (P |S)1. Since P |S is an isomorphism, the vectors (P |S)1(B) form a basis for S.But then

(IdRn |S (P |S)1)(B) = ((P |S + Q|S) (P |S)1)(B)= (IdP +Q|S (P |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 i : Mi Rni be coordinate maps around pi for i = 1, . . . , k,and let : N Hd be a coordinate map around q. Then 1 k : M1 Mk N Hn1++nk+d is a coordinate map, and (1 k )(p1, . . . , pk, q) Hn1++nk+d if and only if (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 1 : (U) Rk is smooth forevery smooth chart (U,) for M .

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

f 1 = f 1 1,which is smooth since 1 is smooth. This shows that f 1 is smooth in aneighborhood of every point in (U), so f 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,) containing p and (V, )containing F (p) such that U F1(V ) is open in M and the composite map F 1 is smooth from (U F1(V )) to (V ).

(2) F is continuous and there exist smooth atlases {(U, )} and {(V, )} for Mand N , respectively, such that for each and , F 1 is a smooth mapfrom (U F1(V)) to (V).

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 and the map F 1 is smooth since for all p M we can findsmooth coordinate maps containing p and containing F (p) such that

F 1 = 1 F 1 1on a small neighborhood around (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,) containing p and (V, ) containing F (p) such that F (U) V and F 1 is smooth. By restricting 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, ) be asmooth chart containing y. If x M and (U,) is any smooth chart containing x then c 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,) is any smooth chart in U containing p then 1 = Id(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 f1 g1 is a smoothinverse to g f , which proves (1). If fi : Mi Ni are diffeomorphisms for i = 1, . . . , nthen f11 f1n is a smooth inverse to f1 fn, which proves (2). Part (3) followsfrom 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,) containing x and (V, ) containing F (x) such that (x) Hn, F (U) = Vand F 1 is a diffeomorphism. Then F1 is a smooth coordinate map sendingF (x) to (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,) containing x and(V, ) containing f(x) such that f 1 is smooth as a map from (U f1(V ))to (V ), but f is not smooth as defined in this chapter.

f is clearly smooth on R\{0}. If x = 0 then = Id(1,1) is a coordinate map containingx and = Id(1/2,3/2) is a coordinate map containing f(x) = 1 such that f 1 issmooth on ((1, 1) f1 ((1/2, 3/2))) = ([0, 1)) = [0, 1). However, f is clearly notsmooth since it is not continuous at 0.

Theorem 18. Let M1, . . . ,Mk1 be smooth manifolds without boundary and let Mk bea smooth manifold with boundary. Then each projection map i : M1 Mk Miis smooth.

Proof. Let ni be the dimension of Mi. Let x = (x1, . . . , xk) M1 Mk. For eachi = 1, . . . , k, choose a smooth chart (Ui, i) containing xi. Write U = U1 Uk and = 1 k. Then (U,) is a smooth chart containing x and i i 1 = piis smooth, where pi is a projection from Rn1 Hnk to Rni (or Hni if i = k). Thisshows that i 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 i : 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 = i 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, i) containing xand (Vi, i) containing Fi(x) such that i Fi 1i = i i F 1i . Replace U1 with

6

U = U1 Uk and replace 1 with 1|U ; by Theorem 12, each map i i F 11is smooth. Write V = V1 Vk and = 1 k so that (V, ) is a smoothchart containing F (x). Since the ith component of F 11 is just i i F 11 ,the map F 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 {f} from C(M) with nonempty disjointsupports is linearly independent. Suppose there exist r1, . . . , rk R such that

r1f1 + + rkfk = 0where 1, . . . , k are distinct. For each i, choose some xi supp fi such that fi(xi) 6=0. Then

rifi(xi) = (r1f1 + + rkfk)(xi) = 0,so ri = 0 since fi(xi) 6= 0.Choose some coordinate cube U M and a homeomorphism : U (0, 1)n. For eachi = 1, 2, . . . , let Bi = (0, 1)

n1 (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

1(Ai)supported in 1(Bi). Furthermore, since the sets {1(Bi)} are disjoint, the functions{fi} 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

F1[y1, . . . , yn, yn+1] = (y1/yn+1, . . . , yn/yn+1);

it is easy to check that the two maps are smooth since F1 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), H1) is a smooth chart for CP1. We can compute

(H1 p H)(z) = (H1 p)[1, z]

=

{H1[p(1/z), 1], z 6= 0,H1[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,) containingx and (V, ) containing F (x) such that F (U) V . For each i = 1, . . . , n the mapi F = F (i ) is smooth, where i : Rn R is the projection onto the ithcoordinate. By 9, i F 1 is smooth for i = 1, . . . , n, so F 1 must besmooth. This shows that F is smooth. Part (3) follows directly from part (2): if Fis a diffeomorphism then (F1) : C(M) C(N) is clearly an inverse to F , andconversely if F |C(N) is an isomorphism then F and F1 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 : 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 : RPn1 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 ,f1(0) = A, and f1(1) = B.

Proof. By Theorem 2.29, there are functions fA, fB : M R such that f1A ({0}) = Aand f1B ({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(F1)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(F1)F (p) dFp = d(F1 F )p = d(IdM)p = IdTpM

and similarly dFp d(F1)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,) containing p and (V, ) containing F (p) such that (U) is an open

ball and F (U) V . Let F = F 1 be the coordinate representation of F . Fromequation (3.10), the derivative of F is zero since dFp = 0 and therefore F is constant

on (U). This implies that F = 1 F 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 j : M1 Mk Mj be the projection onto the Mj factor. For any pointp = (p1, . . . , pk) M1 Mk, the map

: Tp(M1 Mk) Tp1M1 TpkMkdefined by

(v) = (d(1)p(v), . . . , d(k)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 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

: Tp1M1 TpkMk Tp(M1 Mk)(v1, . . . , vk) 7 d(e1)p1(v1) + + d(ek)pk(vk).

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

d(j)p

(ki=1

d(ei)pi(vi)

)=

ki=1

d(j ei)pi(vi)

= d(j ej)pj(vj)= vj

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

dim(Tp(M1 Mk)) = dim(Tp1M1 TpkMk),so is an isomorphism.

Theorem 33. If (Uj, j) are smooth charts containing pj as defined in Theorem 32,then = 1 k is a smooth coordinate map containing p. Let nj be the dimensionof Mj, and write j = (x

1j , . . . , x

njj ) for each j so that

= 1 k = (x11, . . . , xn11 , . . . , x

1k, . . . , x

nkk )

where xij = xij j.

(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 (v) is

i vji /xij

pj

(the variable j is fixed).

Proof. Since Bj ={/x1j

pj, . . . , /x

njj

pj

}is a basis for TpjMj, the set

1(B1 Bk) = (B1 Bk) is a basis for Tp(M1 Mk). But

(

xij

pj

)(f) = d(ej)pj

(

xij

pj

)(f)

= d(ej 1j )j(pj)

(

xij

j(pj)

)(f)

=(f ej 1j )

xij(j(pj))

=(f 1)

xij((p))

12

= d(1)(p)

(

xij

(p)

)(f)

=

xij

p

(f),

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

d(j)p

(r,s

vsr

xrs

p

)(f) =

r,s

vsrd(j)p

(

xrs

p

)(f)

=r

vjr(f j 1)

xrj((p))

=r

vjr(f 1j )

xrj(j(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 (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 (p,q)(v). For any (p, q) M N , choose smooth charts (U,)containing p and (V, ) containing q. Let : T (M N) M N be the naturalprojection map. Then = is a smooth coordinate map for M N , : 1(U V ) R2(m+n) is a smooth coordinate map for T (MN), and : 1(U)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, F1 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 : U R be any angle function defined on a neighborhood of

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z, and set vz = d/dx|z = (d)1z d/dx|(z). If : U R is any other angle functiondefined around z then is constant on U U and ( 1)(t) = t + c for somec R. Therefore (d)1z d/dx|(z) = (d)1z d/dx|(z), and vz is well-defined.

Define a map F : S1 R TS1 by sending (z, r) to rvz. If : U R is an anglefunction defined around z then the coordinate representation of F is F (z, r) = ((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 F1 is F1(x, r) =(1(x), r), so F1 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 : (1, 1) S1 be given by s 7 exp(2i(s+ c)), wherec is chosen so that (F )(0) = (1, 1). We can assume without loss of generality that(F )((, 0)) lies on the right edge of K and that (F )((0, )) lies on the top edge ofK, for small . Then there are smooth functions 1, 2 such that (F )(t) = (1, 1(t))for t (, 0) and (F )(t) = (2(t), 1) for t (0, ). Since (F )(t) = (0, 1(t))and (F )(t) = (2(t), 0), taking t 0 shows that (F )(0) = 0. But dF ((0)) 6= 0since (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 Cp (M) denote the algebra of germs of smooth real-valuedfunctions at p, and let DpM denote the vector space of derivations of Cp (M). Definea map : DpM TpM by (v)f = v([f ]p). Then is an isomorphism.

Proof. It is clear that is a homomorphism. If v = 0 then v([f ]p) = 0 for allf C(M), so v = 0 and therefore 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. Thenv = v, so 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 1 2 if (f 1)(0) = (f 2)(0) for every smooth real-valued function f defined in a neighborhood of p. The map : VpM TpM defined by[] = (0) is well-defined and bijective.

14

Proof. If 1, 2 [] then 1(0)(f) = (f 1)(0) = (f 2)(0) = 2(0)(f), whichshows that is well-defined. The same argument shows that is injective. Finally,Proposition 3.23 shows that 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 )|F1(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 F1(y) and V of y suchthat F |U : U V is a diffeomorphism, so (F1)|V is smooth. Theorem 11 then showsthat F1 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|G1(U) =(F |U)1 (F G)|G1(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, and1 : M N1 and 2 : M N2 are surjective submersions that are constant oneach others fibers. Then there exists a unique diffeomorphism F : N1 N2 such thatF 1 = 2.

Proof. Theorem 4.30 shows that there are unique smooth maps 1 : N2 N1 and2 : N1 N2 such that 1 = 1 2 and 2 = 2 1. Then 2 1 2 = 2 1 = 2,so 2 1 = IdN2 by the uniqueness part of Theorem 4.30. Similarly, 1 2 = 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 : E M be a smooth covering map. Let e E and x = (e), let U be anevenly covered neighborhood of x, and let U be the component of 1(U) containing

16

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

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

Proof. Let : V E be a local section of 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 : U E such that (x) = (x). Therefore |U = ,and is smooth by Theorem 11.

Theorem 47. [Exercise 4.38] Suppose E1, . . . , Ek and M1, . . . ,Mk are smooth man-ifolds (without boundary), and i : Ei Mi is a smooth covering map for eachi = 1, . . . , k. Then 1 k : 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 : E M is a smooth covering map. If U M is evenly cov-ered in the topological sense, then maps each component of 1(U) diffeomorphicallyonto U .

Proof. Let U be a component of 1(U) and suppose |U : U U is a homeomorphism(or merely a bijection). Since |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 : E M is a topological covering map. Then E is a topological n-manifoldwith boundary such that E = 1(M), and it has a unique smooth structure suchthat 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 : M M . The universal covering manifold is unique in thefollowing sense: if M is any other simply connected smooth manifold that admits a

smooth covering map : M M , then there exists a diffeomorphism : M M such that = .

17

Proof. Every manifold has a universal covering space, so let : M M be a universalcovering. Proposition 4.40 shows that M is a smooth manifold with a unique smooth

structure such that is a smooth covering map. If : M M is another universalsmooth covering, then there is a homeomorphism : M M such that =. Let e M , let x = (e) and choose some e 1({x}). Let U be an evenlycovered neighborhood of x with respect to both and , let U be the component of

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

Example 51. [Problem 4-1] Use the inclusion map : 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 d0 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 |U V is adiffeomorphism. Since 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,) centered at p and (V, ) centered at F (p)such that (V ) Hn, F (U) V and (F (p)) = 0. Let F = F 1. Then dF0is 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 aneighborhood of 0 open in Rn and contained in Hn, which is impossible.

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

Let (z1, z2) T2 and choose s1, s2 R so that e2is1 = z1 and e2is2 = z2. Then(s1 + n) = (e

2is1 , e2is1e2in), so it remains to show that

S ={e2in : n Z

}is dense in S1 whenever is irrational. Let > 0. By Lemma 4.21, there exist integersn,m such that |nm| < . Let = nm so that

e2i = e2in Sand 6= 0 since is irrational. Any integer power of e2i is also a member of S, andsince was arbitrary, we can approximate any point on S1 by taking powers of e2i.

18

Example 54. We identify S3 with the set{

(z, w) C2 : |z|2 + |w|2 = 1}

. Let p : CS2 \ {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 zw

), 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 : 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 given by

[w1, . . . , wn, wn+1] 7(

w1wn+1

, . . . ,wnwn+1

).

Then

( )(w1, . . . , wn+1) =

w1n+1 0 0 w1w2n+1

0 w1n+1 0 w2w2n+1...

.... . .

......

0 0 w1n+1 wnw2n+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 : C2\{0} CP1 is a surjectivesmooth submersion. Define a map : C2 \ {0} S2 by

(z1, z2) = (p q)(

(z1, z2)

|(z1, z2)|

),

where p q is the map in 54. By Theorem 4.30, descends to a unique smooth map : CP1 S2 satisfying = . 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 : 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 issmooth if and only if F is smooth, then IdN is a diffeomorphism between N and N .

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

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

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

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

D(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 : E Mbe a topological covering map. There is only one smooth structure on E such that 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 is a smooth covering map. Let p E and let(U, ) be a smooth chart inM containing p. Let V be an evenly covered neighborhoodof (p) and let V be the component of 1(V ) containing p so that |V : V V isa diffeomorphism (with respect to M). By shrinking V and V if necessary, we mayassume that there is a smooth chart (V, ) containing (p). Then |V is a smoothcoordinate map in M, and

( |V )1 = (|V )

1 1

is smooth. This shows that and |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} ,Si = {(x1, . . . , xn+1) Sn : xi < 0} .

Then q|S+i and q|Si are diffeomorphisms for i = 1, . . . , n + 1, which shows that q is alocal 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), themap F descends to a unique smooth map F : RP2 R4. It is easy to check that F1is smooth and given by

F1(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

: RP2 \{

[x, y, z] RP2 : z 6= 0} R2

[x, y, z] 7(xz,y

z

)be coordinate map. Then

(F 1)(x, y) = F [x, y, 1]= (x2 y2, xy, x, y)

so that

D(F 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

(, , ) 7 ( sin cos , sin sin , cos).For the image to be a subset of S2, we must have = 1.Theorem 63. [Exercise 5.24] Suppose M is a smooth manifold with or without bound-ary, S M is an immersed k-submanifold, : 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, ) for S such that X = 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 X1 : X(U) U is a diffeo-morphism, so (X(U), X1) is a smooth chart such that X = (X1)1. Conversely,suppose that (V, ) is a smooth chart such that X = 1. Then X((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 : J M whose image is contained inS, and which is also smooth as a map into S, such that 0 J , (0) = p, and (0) = v.

Proof. Let : S M be the inclusion map. If v TpS then by Proposition 3.23 thereis a smooth curve 0 : J S such that 0 J , 0(0) = p and 0(0) = v. Therefore = 0 is the desired smooth curve. Conversely, if there is a smooth curve withthe specified properties then 0 : J S given by t 7 (t) is smooth. Therefore0(0) TpS and v = (0) = dp(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 : M N with constant rank. Then TpS = ker dp 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 TpM , 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, . . . , n1 and (f/xn)(p) > 0.Write v =

ni=1 v

i(/xi|p) so that

vf =ni=1

vif

xi(p) = vn

f

xn(p);

then the result follows from Theorem 66.

Example 68. [Problem 5-1] Consider the map : R4 R2 defined by(x, y, s, t) = (x2 + y, x2 + y2 + s2 + t2 + y).

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

We compute

D(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 (x, y, s, t) =(0, 1). This shows that dp is surjective for all p S. Define : R4 R4 by

(x, y, s, t) = (x, x2 + y, s, t);

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

(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 |S is a diffeomorphism. Therefore S is diffeomorphic to (S).

Theorem 69. [Problem 5-2] If M is a smooth n-manifold with boundary, then with thesubspace topology, M is a topological (n1)-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,) containing x. Then

(M U) ={

(x1, . . . , xn) (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 : S M is anopen map. The result follows from Proposition 4.22.

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

(t) = (sin 2t, sin t)

is not an embedded submanifold of R2.

It is well-known that the image of 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 : R T2 be the curve of Example 4.20. Show that(R) is not an embedded submanifold of the torus.

Again, (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 (2m1)-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,) for Rn containing x such that

(M U) ={

(x1, . . . , xn) (U) : xm+1 = = xn = 0}.

24

Similarly, Sm1 is an embedded submanifold of Rm, so we can choose a smooth chart(V, ) for Rm containing v such that

(Sm1 V ) ={

(x1, . . . , xm) (V ) : xm = 0}.

Write = (x1, . . . , xn) and U = 1(U). By definition, the map : 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 ((U)), where : R2n Rm is the projection onto the coordinates n+ 1, . . . , n+m. Define

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

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

( )(UM U)

={

(x1, . . . , xn, v1, . . . , vn) ( )(U) : xm+1 = = xn = vm = = vn = 0},

so UM satisfies the local (2m1)-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 F1({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 = {0}n1 [0, 1] be the vertical line connecting 0 and N . Define : Bn \ T Hn given by

(x1, . . . , xn) =

(x1

x xn, . . . ,

xn1x xn

, x1 1),

with its inverse given by

1(u1, . . . , un) =1

(un + 1)(u2 + 1)(2u1, . . . , 2un1, u2 1)

where u = (u1, . . . , un1). 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 = {0}n1 [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 Sn1.

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 map : S R2 is a smooth immersion. By Theorem 4.25, there is a neighborhood Uof (1, 1) such that |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 byMa =

{(x, y) : y2 = x(x 1)(x a)

}.

For which values of a is Ma an embedded submanifold of R2? For which values can Mabe 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 : E M is a smooth covering map. The restriction of 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 (F ), let U be an evenlycovered neighborhood of x, choose e so that x = (e), and let U be the component

of 1(U) containing e. Then |U : U U is a diffeomorphism. It is clear that(U F ) = U M . Since U F is an embedded submanifold of U and (U F ) is anembedded submanifold of U by Theorem 5.11, |UF is a diffeomorphism. This showsthat U M is an evenly covered neighborhood of x, and that the image of |F is open.Let x / (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 = (f) V . Let U be the componentof 1(U) containing f and choose e U so that x = (e). Let be a path from xto y in V . Then (|U)1 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 / (F ), soV M \ (F ). This shows that (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 : S M be the inclusion map, which is a smooth immersion. Let S denoteS with some other smooth structure such that : S M is a smooth immersion.Since S and S have the same topology, the maps : S S and : S S areboth continuous, and Theorem 5.29 shows that the maps are inverses of each other.Therefore S is diffeomorphic to S .

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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, p) for S in M containing p. Then SUp is closed in Up, so Proposition 2.25 showsthat there is smooth function fp : Up R such that fp|SUp = f |SUp . Let {p}pS bea partition of unity subordinate to {Up}pS; then

f(x) =pS

p(x)fp(x)

is a smooth function defined on U =pS 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 havea partition of unity {p} {0} 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 : J M is a smooth curve whose image happens to lie in S. Then (t) is in the subspaceT(t)S of T(t)M for all t J . This need not be true if S is not embedded.

Proof. Corollary 5.30 shows that is smooth as a map into S. Denote this map by0 : J S and let : S M . Then = 0, so (t) = d(t)(0(t)) by Proposition3.24 and therefore (t) T(t)S T(t)M . This need not hold if S is merely animmersed submanifold. For example, consider a curve that crosses the point of self-intersection in the figure-eight curve of Example 4.19, such that is not continuousas a map into the image of .

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 f1 ((, b]) is a regular domainin M .

(2) If a and b are two regular values of f with a < b, then f1 ([a, b]) is a regulardomain in M .

28

Proof. Let m be the dimension of M . It is clear that f1 ((, b)) satisfies the localm-slice condition. Furthermore, there is a boundary slice chart around each pointof f1({b}). We can see this by applying the argument in Corollary 5.14, with thefollowing modification to Theorem 5.12: for each p f1({b}), the rank theorem showsthat there are smooth charts (U,) centered at p and (V, ) containing b for which the

coordinate representation f of f has the form

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

By modifying f and appropriately, U becomes a boundary slice chart for f1 ((, b])in M . Theorem 5.51 then shows that f1 ((, 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, p)}pD be a collection of interior or boundary slice charts for Din M . Then U {M \D} covers M , and we can repeat the argument of Proposition5.43 by constructing functions {fp}pD {f0} as follows: set fp = 1 when p IntD,set fp(x

1, . . . , xn) = xn when p D, and set f0 = 1 for f0 : M \ D R. If D iscompact 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. Sards 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,) 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

(n=1

An U

)=n=1

(An U)

has measure zero (in Rn) since each (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 composition (F IdRk), where : N Rk N is the canonical projection. It suffices to showthat the image of F has measure zero in N . Let (U1U2, 12) and (V, ) be smoothcharts for MRk and N respectively such that U1 M and U2 Rk. Let U = U1U2and = 1 2. Then

(F (U) V ) = (F (U F1(V )))

= ( F 1)((U F1(V )))

( F 1)((U) (1(F1(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, soF (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 smoothbecause it is the composition UM R2n+1 \ {0} RP2n. Since dimUM < dimRP2n,Sards theorem shows that the image of G has measure zero in RP2n, and since (R2n)has measure zero in RP2n, there is a v R2n+1 \ R2n such that [v] is not in the imageof G. Let v : R2n+1 R2n be the projection with kernel Rv; then v|M is smooth.Furthermore, if p M then v / TpM , so d(v|M)p is injective.

Theorem 89. [Problem 6-3] Let M be a smooth manifold, let B M be a closedsubset, and let : M R be a positive continuous function. There is a smoothfunction : M R that is zero on B, positive on M \ B, and satisfies (x) < (x)everywhere.

Proof. By Theorem 2.29, there is a smooth nonnegative function f : M R such thatf1({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 that0 < e(x) < (x) for all x M . Then = 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 properin Theorem 6.21, then F is also proper. Let E Rk be compact. Since is continuous,it attains some maximum (x0) on E. Let B be a closed ball containing E such thatfor every x E, the closed ball of radius (x) around x is contained in B. Since|F (x) F (x)| < (x), it is clear that F1(E) is a closed subset of F1(B). But F isproper, so F1(B) is compact and therefore F1(E) is compact.

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

F (x, y) = (ey cosx, ey sinx, ey).

For which positive numbers r is F transverse to the sphere Sr(0) R3? For whichpositive numbers r is F1 (Sr(0)) an embedded submanifold of R2?

We compute

DF (x, y) =

ey sinx ey cosxey cosx ey sinx0 ey

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

r2 = e2y cos2 x+ e2y sin2 x+ e2y = e2y + e2y,

i.e.

y =1

2log

(1

2(r2

r4 4)

).

Thus F1 (Sr(0)) is always an embedded submanifold of R2, being either empty (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 sinx0 1

.

31

The corresponding point in S2(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)S2(0) is a proper subset of Tv(x)R3.

Theorem 93. [Problem 6-15] Suppose M and N are smooth manifolds and S MNis an immersed submanifold. Let M and N 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) M |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 =N (M |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 M |S is smooth with inverse g. Conversely, if M |S is a diffeomorphismthen we can take f = N (M |S)1 in (1). This proves that (1) (2). If (1) holdsthen d(M)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 M |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(M |S)p(TpS) m.Therefore d(M |S)p is invertible for every p M , and M |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 : GG G given by (g, h) 7 gh1 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 : GGG is a smooth submersion.

Proof. For each g G, the map g : G G G given by x 7 (g, g1x) is a smoothlocal section of m since m(g(x)) = gg

1x = x. Every point (g1, g2) GG is in theimage of g1 , 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)(GG) with TeGTeG, the differential dm(e,e) : TeGTeG 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 bydie(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 : GG 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 :GG GG 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) GGthen it is easy to check that

F = ((Lxy Ry1) IdG) F ((Ly1x1 Ry) Ly1),so dF(x,y) = dF(e,e) where , 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 F1(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(X1B).

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 + tX1B),

so

d(det)X(B) =d

dt

t=0

det(X + tB)

= det(X)d

dt

t=0

det(In + tX1B)

= det(X) tr(X1B).

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 : G G and : G G that are also Lie grouphomomorphisms, then there exists a Lie group isomorphism : G G such that = .

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

is a diffeomorphism : G G such that = , so it remains to show that is agroup homomorphism. By replacing with (e)1 = L(e)1 , we can assume that(e) = e. (Since ((e)1(g)) = ((e))1(g) = (g), we still have = .) Ifg, h G then

((gh)) = (g)(h) = ((g)(h)),

so (g, h) 7 (gh) and (g, h) 7 (g)(h) are both lifts of (g, h) 7 (gh). Since themaps all agree at the point (e, e), we have (gh) = (g)(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 andgh1 U whenever g, h V .

Proof. Define f(g, h) = gh1 and let W = f1(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 G0since G0 is a connected component. Therefore H = G0. For each g G, define theconjugation map Cg : G G by h 7 ghg1. 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 : G K K be the group action. For any x K the set (G {x}) isconnected and therefore consists of one point. But (e, x) = x, so (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 : 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 map : GL(n + 1,R) RPn RPn given by (A, [x]) 7 [Ax]. It is easy to see that is agroup action. Furthermore, for any two points v, w Rn+1 there is an invertible linearmap that takes v to w, so 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 sin sin cos

],

where = 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 r1

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) 7nh is a Lie group isomorphism between N o H and G, where : H N N is theaction by conjugation: h(n) = hnh

1. Furthermore, N and H are closed in G.

36

Proof. Let be the map (n, h) 7 nh. If nh = e then n = h1 N H, so n = h = e.Also, is surjective since NH = G. Therefore is a bijection. We have

((n, h)(n, h)) = (nhnh1, hh)

= nhnh1hh

= nhnh

= (n, h)(n, h),

so is an isomorphism. Since 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 homomorphisms : G H and : H G such that = IdH and ker = N .

Proof. Suppose there is an isomorphism f : G N o H. Let : N o H Hbe the projection (n, h) 7 h and let : H N o H be the injection h 7 (e, h).Take = f and = ; then = IdH and ker = N . Conversely, suppose thehomomorphisms and exist. Let N = ker and H1 = (H). Then N is normal in Gand N H1 = {e}, since x = (h) and x ker implies that h = ((h)) = (x) = e.For any g G we have (((g))) = (g), so g = n((g)) for some n ker.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 by M 7 det(M) and by

z 7[z

In1

]where In1 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 XhXhgY 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

[ni=1

xiEi,

nj=1

yjEj

]

=ni=1

xi

nj=1

yjA [Ei, Ej]

=ni=1

xi

nj=1

yj [AEi, AEj]

38

=

[A

ni=1

xiEi, A

nj=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 : 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 : Lie(GL(V )) Lie(GL(n,R)) induced by the diffeomorphism , and by Proposition 8.41 there is a Liealgebra isomorphism : Lie(GL(n,R)) gl(n,R). Then 1 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. Givenany 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 Xpon 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 . Let{p : p A} {0} be a smooth partition of unity subordinate to this cover, withsuppp Wp and supp0 M \ A.

For each p A, the product pXp 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\suppp. (The extended functionis smooth because the two definitions agree on the open subset Wp \ suppp where theyoverlap.) Thus we can define X : M TM by

Xx =pA

p(x)Xp|x.

Because the collection of supports {suppp} 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 0(x) = 0 and Xp|x = Xx for each p suchthat p(x) 6= 0, so

Xx =pA

p(x)Xx =

(0(x) +

pA

p(x)

)Xx = Xx,

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

supp X =pA

suppp =pA

suppp 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(x) = cf(x) for all > 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 =ni=1

xif

xi(x).

Differentiating cf(x) = f(x) with respect to on (0,), we have

cc1f(x) = Df(x)(x) =ni=1

xif

xi(x).

Replacing x with 1x gives

cc1f(1x) = 1ni=1

xif

xi(x)

cc1cf(x) = 1ni=1

xif

xi(x)

cf(x) =ni=1

xif

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 : (, 1 + ) U where > 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 (, 1 + ) by Lemma 8.6. We willshow that {Xm : m = 1, 2, . . . } is linearly independent by induction. Since Xm isnonzero, 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{Xm : m = 1, 2, . . . } is a set of linearly independent vector fields defined on U . Wecan extend these vector fields to M by restricting each Xm to the closed set ([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 {(U, )} be a collection of smooth boundary charts whosedomains cover M . For each , let X be the nth coordinate vector field. Let {}be a partition of unity of U =

U subordinate to {U}; then X =

X is a

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

Xpf =

X|pf > 0

since each X|p is inward-pointing, so Xp is always inward-pointing. By restricting X toM 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) onsome 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

fehg

=(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 : [0, 1) S1 given by s 7 e2is 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 FX and FY , where

X = x

x+ y

y, Y = y

x.

The inverse

F1(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 (F1(u, v)) =

[ uv

u/v

vv/u

v/u

].

Therefore the coordinates of FX are given by[ uv

u/v

vv/u

v/u

] [u/vuv

]=

[2u0

],

43

and the coordinates of FY are given by[ uv

u/v

vv/u

v/u

] [uv0

]=

[uvv2

].

So

FX = 2u

u, FY = 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,) (, )given by F (x, y) = (

x2 + y2, arctan(y/x)), with inverse F1(r, ) = (r cos , r sin ).

Then

DF (x, y) =

[x/x2 + y2 y/

x2 + y2

y/(x2 + y2) x/(x2 + y2)

],

so

DF (F1(r, )) =

[cos sin

r1 sin r1 cos

].

We have[cos sin

r1 sin r1 cos

] [r cos r cos r2

r sin r sin 0

]=

[r r cos 2 r2 cos 0 sin 2 r sin

].

Therefore

X = r

r,

Y = r cos 2

r sin 2

,

Z = r2 cos

r r sin

.

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 : R3 \ {0} RP2 be thequotient map and define X : R3 \ {0} TRP2 by

X(x,y,z) = d

(x

y

(x,y,z)

y x

(x,y,z)

).

Since X is constant on the fibers of , it descends to a vector field Y : RP2 TRP2.Let Ui R3 \ {0}, Ui RP2 and i : Ui Rn be as in Example 1.5:

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

1i (u1, u2, u3) = [u1, . . . , ui1, 1, ui+1, . . . , u3].

The coordinate representations of d and Y with respect to 3 are given by

d(u1, u2, u3, v1, v2, v3) =

(u1

u3,u2

u3, 1,

v1

u3 u

1v3

(u3)2,v2

u3 u

2v3

(u3)2, 0

)and

Y (u1, u2, u3) = d

(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 1 and 2 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 : S2\{N} R2 be the stereographicprojection given by

(x1, x2, x3) = (x1, x2)/(1 x3),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,

((1)X)p p 6= N.

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

(x1, x2, x3) = (x1, x2)/(1 + x3),

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

( 1)(u1, u2) = (2u1, 2u2)/(2 |u|2),and the coordinate representation of Y for (u1, u2) 6= (0, 0) with respect to is givenby

Y (u1, u2) = d(1)(1)(u1,u2)X(1)(u1,u2)

= d(1)(1)(u1,u2)

(d

dx

(1)(u1,u2)

)

=

(1(u1, u2),

2(1 (u1)2 + (u2)2)(1 + |u|2)2

, 4u1u2

(1 + |u|2)2, 4u

1

(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 MN . Let p F (M), let (U,) be a slice chart for F (M)containing p, let p = (p), and let U = (U) Rm+n. By shrinking U , we may assumethat U is an open cube containing p, and that (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 d(X(1(a,0))), where : MN M is the canonical projection. ThenY is smooth, and (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 haveX2

x= 2y2, X

2

y= 4xy, X

3

y= 1,

so

[X, Y ] = 4xy

y z.

For (2), we have

X1

y= 1, X

2

x= 1,

Y 2

z= 1, Y

3

y= 1,

so

[X, Y ] = z

x x

z.

For (3), we have

X1

y= 1, X

2

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 XY follows easily from Proposition 8.1. If f C(MN)then

[X1 Y1, X2 Y2](