the following property is true:

(1) For each x ∈ M there exists a chart (U, u) centered at x on N such that

u(Cx (U © M )) = u(U ) © (Rm — 0).

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

2. Submersions and immersions 13

The following three lemmas explain the name initial submanifold.

2.15. Lemma. Let f : M ’ N be an injective immersion between manifolds

with property 2.9.1. Then f (M ) is an initial submanifold of N .

Proof. Let x ∈ M . By 2.6 we may choose a chart (V, v) centered at f (x) on N

and another chart (W, w) centered at x on M such that (v—¦f —¦w’1 )(y 1 , . . . , y m ) =

(y 1 , . . . , y m , 0, . . . , 0). Let r > 0 be so small that {y ∈ Rm : |y| < r} ‚ w(W )

and {z ∈ Rn : |z| < 2r} ‚ v(V ). Put

U : = v ’1 ({z ∈ Rn : |z| < r}) ‚ N,

W1 : = w’1 ({y ∈ Rm : |y| < r}) ‚ M.

We claim that (U, u = v|U ) satis¬es the condition of 2.14.1.

u’1 (u(U ) © (Rm — 0)) = u’1 ({(y 1 , . . . , y m , 0 . . . , 0) : |y| < r}) =

= f —¦ w’1 —¦ (u —¦ f —¦ w’1 )’1 ({(y 1 , . . . , y m , 0 . . . , 0) : |y| < r}) =

= f —¦ w’1 ({y ∈ Rm : |y| < r}) = f (W1 ) ⊆ Cf (x) (U © f (M )),

since f (W1 ) ⊆ U © f (M ) and f (W1 ) is C ∞ -contractible.

Now let conversely z ∈ Cf (x) (U ©f (M )). Then by de¬nition there is a smooth

curve c : [0, 1] ’ N with c(0) = f (x), c(1) = z, and c([0, 1]) ⊆ U © f (M ). By

property 2.9.1 the unique curve c : [0, 1] ’ M with f —¦ c = c, is smooth.

¯ ¯

We claim that c([0, 1]) ⊆ W1 . If not then there is some t ∈ [0, 1] with c(t) ∈

¯ ¯

’1 m

w ({y ∈ R : r ¤ |y| < 2r}) since c is smooth and thus continuous. But then

¯

we have

(v —¦ f )(¯(t)) ∈ (v —¦ f —¦ w’1 )({y ∈ Rm : r ¤ |y| < 2r}) =

c

= {(y, 0) ∈ Rm — 0 : r ¤ |y| < 2r} ⊆ {z ∈ Rn : r ¤ |z| < 2r}.

This means (v —¦ f —¦ c)(t) = (v —¦ c)(t) ∈ {z ∈ Rn : r ¤ |z| < 2r}, so c(t) ∈ U , a

¯ /

contradiction.

So c([0, 1]) ⊆ W1 , thus c(1) = f ’1 (z) ∈ W1 and z ∈ f (W1 ). Consequently we

¯ ¯

have Cf (x) (U © f (M )) = f (W1 ) and ¬nally f (W1 ) = u’1 (u(U ) © (Rm — 0)) by

the ¬rst part of the proof.

2.16. Lemma. Let M be an initial submanifold of a manifold N . Then there

is a unique C ∞ -manifold structure on M such that the injection i : M ’ N

is an injective immersion. The connected components of M are separable (but

there may be uncountably many of them).

Proof. We use the sets Cx (Ux © M ) as charts for M , where x ∈ M and (Ux , ux )

is a chart for N centered at x with the property required in 2.14.1. Then the

chart changings are smooth since they are just restrictions of the chart changings

on N . But the sets Cx (Ux © M ) are not open in the induced topology on M

in general. So the identi¬cation topology with respect to the charts (Cx (Ux ©

M ), ux )x∈M yields a topology on M which is ¬ner than the induced topology, so

it is Hausdor¬. Clearly i : M ’ N is then an injective immersion. Uniqueness of

the smooth structure follows from the universal property of lemma 2.17 below.

Finally note that N admits a Riemannian metric since it is separable, which can

be induced on M , so each connected component of M is separable.

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

14 Chapter I. Manifolds and Lie groups

2.17. Lemma. Any initial submanifold M of a manifold N with injective

immersion i : M ’ N has the universal property 2.9.1:

For any manifold Z a mapping f : Z ’ M is smooth if and only if i —¦ f : Z ’

N is smooth.

Proof. We have to prove only one direction and we will suppress the embedding i.

For z ∈ Z we choose a chart (U, u) on N , centered at f (z), such that u(Cf (z) (U ©

M )) = u(U ) © (Rm — 0). Then f ’1 (U ) is open in Z and contains a chart (V, v)

centered at z on Z with v(V ) a ball. Then f (V ) is C ∞ -contractible in U © M , so

f (V ) ⊆ Cf (z) (U ©M ), and (u|Cf (z) (U ©M ))—¦f —¦v ’1 = u —¦f —¦v ’1 is smooth.

2.18. Transversal mappings. Let M1 , M2 , and N be manifolds and let

fi : Mi ’ N be smooth mappings for i = 1, 2. We say that f1 and f2 are

transversal at y ∈ N , if

im Tx1 f1 + im Tx2 f2 = Ty N whenever f1 (x1 ) = f2 (x2 ) = y.

Note that they are transversal at any y which is not in f1 (M1 ) or not in f2 (M2 ).

The mappings f1 and f2 are simply said to be transversal, if they are transversal

at every y ∈ N .

If P is an initial submanifold of N with injective immersion i : P ’ N , then

f : M ’ N is said to be transversal to P , if i and f are transversal.

Lemma. In this case f ’1 (P ) is an initial submanifold of M with the same

codimension in M as P has in N , or the empty set. If P is a submanifold, then

also f ’1 (P ) is a submanifold.

Proof. Let x ∈ f ’1 (P ) and let (U, u) be an initial submanifold chart for P

centered at f (x) on N , i.e. u(Cx (U © P )) = u(U ) © (Rp — 0). Then the mapping

f pr2

u

M ⊇ f ’1 (U ) ’ U ’ u(U ) ⊆ Rp — Rn’p ’ ’ Rn’p

’ ’ ’

is a submersion at x since f is transversal to P . So by lemma 2.2 there is a chart

(V, v) on M centered at x such that we have

(pr2 —¦ u —¦ f —¦ v ’1 )(y 1 , . . . , y n’p , . . . , y m ) = (y 1 , . . . , y n’p ).

But then z ∈ Cx (f ’1 (P ) © V ) if and only if v(z) ∈ v(V ) © (0 — Rm’n+p ), so

v(Cx (f ’1 (P ) © V )) = v(V ) © (0 — Rm’n+p ).

2.19. Corollary. If f1 : M1 ’ N and f2 : M2 ’ N are smooth and transver-

sal, then the topological pullback

— M2 = M1 —N M2 := {(x1 , x2 ) ∈ M1 — M2 : f1 (x1 ) = f2 (x2 )}

M1

(f1 ,N,f2 )

is a submanifold of M1 — M2 , and it has the following universal property.

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2. Submersions and immersions 15

For any smooth mappings g1 : P ’ M1 and g2 : P ’ M2 with f1 —¦g1 = f2 —¦g2

there is a unique smooth mapping (g1 , g2 ) : P ’ M1 —N M2 with pr1 —¦ (g1 , g2 ) =

g1 and pr2 —¦ (g1 , g2 ) = g2 .

RR g2

P

R ,g )

T

R

(g 1 2

u

wM

M1 —N M2

g1 2

pr2

u u

pr1 f2

wM w

N 1

f1

This is also called the pullback property in the category Mf of smooth man-

ifolds and smooth mappings. So one may say, that transversal pullbacks exist

in the category Mf .

Proof. M1 —N M2 = (f1 — f2 )’1 (∆), where f1 — f2 : M1 — M2 ’ N — N and

where ∆ is the diagonal of N — N , and f1 — f2 is transversal to ∆ if and only if

f1 and f2 are transversal.

2.20. The category of ¬bered manifolds. Consider a ¬bered manifold

(M, p, N ) from 2.4 and a point x ∈ N . Since p is a surjective submersion, the

injection ix : x ’ N of x into N and p : M ’ N are transversal. By 2.19, p’1 (x)

is a submanifold of M , which is called the ¬ber over x ∈ N .

¯¯¯ ¯¯¯

Given another ¬bered manifold (M , p, N ), a morphism (M, p, N ) ’ (M , p, N )

means a smooth map f : M ’ N transforming each ¬ber of M into a ¬ber of

¯ ¯¯ ¯

M . The relation f (Mx ) ‚ Mx de¬nes a map f : N ’ N , which is characterized

by the property p —¦ f = f —¦ p. Since p —¦ f is a smooth map, f is also smooth by

¯ ¯

2.4. Clearly, all ¬bered manifolds and their morphisms form a category, which

will be denoted by FM. Transforming every ¬bered manifold (M, p, N ) into its

¯¯¯

base N and every ¬bered manifold morphism f : (M, p, N ) ’ (M , p, N ) into the

¯

induced map f : N ’ N de¬nes the base functor B : FM ’ Mf .

¯¯

If (M, p, N ) and (M , p, N ) are two ¬bered manifolds over the same base N ,

¯ ¯

then the pullback M —(p,N,p) M = M —N M is called the ¬bered product of M

¯

¯ ¯

and M . If p, p and N are clear from the context, then M —N M is also denoted

¯

¯ ¯¯¯

by M • M . Moreover, if f1 : (M1 , p1 , N ) ’ (M1 , p1 , N ) and f2 : (M2 , p2 , N ) ’

¯¯¯ ¯

(M2 , p2 , N ) are two FM-morphisms over the same base map f0 : N ’ N , then

¯ ¯¯

the values of the restriction f1 — f2 |M1 —N M2 lie in M1 —N M2 . The restricted

¯ ¯¯

map will be denoted by f1 —f0 f2 : M1 —N M2 ’ M1 —N M2 or f1 •f2 : M1 •M2 ’

¯ ¯

M1 • M2 and will be called the ¬bered product of f1 and f2 .

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

16 Chapter I. Manifolds and Lie groups

3. Vector ¬elds and ¬‚ows

3.1. De¬nition. A vector ¬eld X on a manifold M is a smooth section of

the tangent bundle; so X : M ’ T M is smooth and πM —¦ X = IdM . A local

vector ¬eld is a smooth section, which is de¬ned on an open subset only. We

denote the set of all vector ¬elds by X(M ). With point wise addition and scalar

multiplication X(M ) becomes a vector space.

‚ ‚

Example. Let (U, u) be a chart on M . Then the ‚ui : U ’ T M |U , x ’ ‚ui |x ,

described in 1.6, are local vector ¬elds de¬ned on U .

Lemma. If X is a vector ¬eld on M and (U, u) is a chart on M and x ∈ U , then

m m

‚ ‚

we have X(x) = i=1 X(x)(ui ) ‚ui |x . We write X|U = i=1 X(ui ) ‚ui .

‚

3.2. The vector ¬elds ( ‚ui )m on U , where (U, u) is a chart on M , form a

i=1

holonomic frame ¬eld. By a frame ¬eld on some open set V ‚ M we mean

m = dim M vector ¬elds si ∈ X(V ) such that s1 (x), . . . , sm (x) is a linear basis

of Tx M for each x ∈ V . In general, a frame ¬eld on V is said to be holonomic, if

‚

V can be covered by an atlas (U± , u± )±∈A such that si |U± = ‚ui for all ± ∈ A.

±

In the opposite case, the frame ¬eld is called anholonomic.

With the help of partitions of unity and holonomic frame ¬elds one may

construct ˜many™ vector ¬elds on M . In particular the values of a vector ¬eld

can be arbitrarily preassigned on a discrete set {xi } ‚ M .

3.3. Lemma. The space X(M ) of vector ¬elds on M coincides canonically with

the space of all derivations of the algebra C ∞ (M, R) of smooth functions, i.e.

those R-linear operators D : C ∞ (M, R) ’ C ∞ (M, R) with D(f g) = D(f )g +

f D(g).

Proof. Clearly each vector ¬eld X ∈ X(M ) de¬nes a derivation (again called

X, later sometimes called LX ) of the algebra C ∞ (M, R) by the prescription

X(f )(x) := X(x)(f ) = df (X(x)).

If conversely a derivation D of C ∞ (M, R) is given, for any x ∈ M we consider

Dx : C ∞ (M, R) ’ R, Dx (f ) = D(f )(x). Then Dx is a derivation at x of

C ∞ (M, R) in the sense of 1.5, so Dx = Xx for some Xx ∈ Tx M . In this

way we get a section X : M ’ T M . If (U, u) is a chart on M , we have

m i‚

i=1 X(x)(u ) ‚ui |x by 1.6. Choose V open in M , V ‚ V ‚ U , and

Dx =

• ∈ C ∞ (M, R) such that supp(•) ‚ U and •|V = 1. Then • · ui ∈ C ∞ (M, R)

and (•ui )|V = ui |V . So D(•ui )(x) = X(x)(•ui ) = X(x)(ui ) and X|V =

m ‚

i

i=1 D(•u )|V · ‚ui |V is smooth.

3.4. The Lie bracket. By lemma 3.3 we can identify X(M ) with the vector

space of all derivations of the algebra C ∞ (M, R), which we will do without any

notational change in the following.

If X, Y are two vector ¬elds on M , then the mapping f ’ X(Y (f ))’Y (X(f ))

is again a derivation of C ∞ (M, R), as a simple computation shows. Thus there is

a unique vector ¬eld [X, Y ] ∈ X(M ) such that [X, Y ](f ) = X(Y (f )) ’ Y (X(f ))

holds for all f ∈ C ∞ (M, R).

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

3. Vector ¬elds and ¬‚ows 17

‚

X i ‚ui

In a local chart (U, u) on M one immediately veri¬es that for X|U =

‚

and Y |U = Y i ‚ui we have

X i ‚ui , Y j ‚uj = X i ( ‚ui Y j ) ’ Y i ( ‚ui X j )

‚ ‚ ‚ ‚ ‚

‚uj ,

i j i,j

since second partial derivatives commute. The R-bilinear mapping

] : X(M ) — X(M ) ’ X(M )

[,

is called the Lie bracket. Note also that X(M ) is a module over the algebra

C ∞ (M, R) by point wise multiplication (f, X) ’ f X.

] : X(M ) — X(M ) ’ X(M ) has the following

Theorem. The Lie bracket [ ,

properties:

[X, Y ] = ’[Y, X],

[X, [Y, Z]] = [[X, Y ], Z] + [Y, [X, Z]], the Jacobi identity,

[f X, Y ] = f [X, Y ] ’ (Y f )X,

[X, f Y ] = f [X, Y ] + (Xf )Y.

The form of the Jacobi identity we have chosen says that ad(X) = [X, ] is

a derivation for the Lie algebra (X(M ), [ , ]).

The pair (X(M ), [ , ]) is the prototype of a Lie algebra. The concept of a

Lie algebra is one of the most important notions of modern mathematics.

Proof. All these properties can be checked easily for the commutator [X, Y ] =

X —¦ Y ’ Y —¦ X in the space of derivations of the algebra C ∞ (M, R).

3.5. Integral curves. Let c : J ’ M be a smooth curve in a manifold M

de¬ned on an interval J. We will use the following notations: c (t) = c(t) =

™

d

dt c(t) := Tt c.1. Clearly c : J ’ T M is smooth. We call c a vector ¬eld along

c since we have πM —¦ c = c.

A smooth curve c : J ’ M will be called an integral curve or ¬‚ow line of a

vector ¬eld X ∈ X(M ) if c (t) = X(c(t)) holds for all t ∈ J.

3.6. Lemma. Let X be a vector ¬eld on M . Then for any x ∈ M there is

an open interval Jx containing 0 and an integral curve cx : Jx ’ M for X (i.e.

cx = X —¦ cx ) with cx (0) = x. If Jx is maximal, then cx is unique.

Proof. In a chart (U, u) on M with x ∈ U the equation c (t) = X(c(t)) is an

ordinary di¬erential equation with initial condition c(0) = x. Since X is smooth

there is a unique local solution by the theorem of Picard-Lindel¨f, which even

o

depends smoothly on the initial values, [Dieudonn´ I, 69, 10.7.4]. So on M there

e

are always local integral curves. If Jx = (a, b) and limt’b’ cx (t) =: cx (b) exists

in M , there is a unique local solution c1 de¬ned in an open interval containing

b with c1 (b) = cx (b). By uniqueness of the solution on the intersection of the

two intervals, c1 prolongs cx to a larger interval. This may be repeated (also on

the left hand side of Jx ) as long as the limit exists. So if we suppose Jx to be

maximal, Jx either equals R or the integral curve leaves the manifold in ¬nite

(parameter-) time in the past or future or both.

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

18 Chapter I. Manifolds and Lie groups

3.7. The ¬‚ow of a vector ¬eld. Let X ∈ X(M ) be a vector ¬eld. Let us

write FlX (x) = FlX (t, x) := cx (t), where cx : Jx ’ M is the maximally de¬ned

t

integral curve of X with cx (0) = x, constructed in lemma 3.6. The mapping FlX

is called the ¬‚ow of the vector ¬eld X.

Theorem. For each vector ¬eld X on M , the mapping FlX : D(X) ’ M is

smooth, where D(X) = x∈M Jx — {x} is an open neighborhood of 0 — M in

R — M . We have

FlX (t + s, x) = FlX (t, FlX (s, x))

in the following sense. If the right hand side exists, then the left hand side exists

and we have equality. If both t, s ≥ 0 or both are ¤ 0, and if the left hand side

exists, then also the right hand side exists and we have equality.

Proof. As mentioned in the proof of 3.6, FlX (t, x) is smooth in (t, x) for small

t, and if it is de¬ned for (t, x), then it is also de¬ned for (s, y) nearby. These are

local properties which follow from the theory of ordinary di¬erential equations.

Now let us treat the equation FlX (t + s, x) = FlX (t, FlX (s, x)). If the right

hand side exists, then we consider the equation

FlX (t + s, x) = FlX (u, x)|u=t+s = X(FlX (t + s, x)),

d d

dt du

FlX (t + s, x)|t=0 = FlX (s, x).

But the unique solution of this is FlX (t, FlX (s, x)). So the left hand side exists

and equals the right hand side.

If the left hand side exists, let us suppose that t, s ≥ 0. We put

FlX (u, x) if u ¤ s

cx (u) =

FlX (u ’ s, FlX (s, x)) if u ≥ s.

FlX (u, x) = X(FlX (u, x))

d

for u ¤ s

du

d

du cx (u) = =

FlX (u ’ s, FlX (s, x)) = X(FlX (u ’ s, FlX (s, x)))

d

du

for 0 ¤ u ¤ t + s.

= X(cx (u))

Also cx (0) = x and on the overlap both de¬nitions coincide by the ¬rst part of

the proof, thus we conclude that cx (u) = FlX (u, x) for 0 ¤ u ¤ t + s and we

have FlX (t, FlX (s, x)) = cx (t + s) = FlX (t + s, x).

Now we show that D(X) is open and FlX is smooth on D(X). We know

already that D(X) is a neighborhood of 0 — M in R — M and that FlX is smooth

near 0 — M .

For x ∈ M let Jx be the set of all t ∈ R such that FlX is de¬ned and smooth

on an open neighborhood of [0, t] — {x} (respectively on [t, 0] — {x} for t < 0)

in R — M . We claim that Jx = Jx , which ¬nishes the proof. It su¬ces to show

that Jx is not empty, open and closed in Jx . It is open by construction, and

not empty, since 0 ∈ Jx . If Jx is not closed in Jx , let t0 ∈ Jx © (Jx \ Jx ) and

suppose that t0 > 0, say. By the local existence and smoothness FlX exists and is

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

3. Vector ¬elds and ¬‚ows 19

smooth near [’µ, µ] — {y := FlX (t0 , x)} for some µ > 0, and by construction FlX

exists and is smooth near [0, t0 ’ µ] — {x}. Since FlX (’µ, y) = FlX (t0 ’ µ, x) we

conclude for t near [0, t0 ’ µ], x near x, and t near [’µ, µ], that FlX (t + t , x ) =

FlX (t , FlX (t, x )) exists and is smooth. So t0 ∈ Jx , a contradiction.

3.8. Let X ∈ X(M ) be a vector ¬eld. Its ¬‚ow FlX is called global or complete,

if its domain of de¬nition D(X) equals R — M . Then the vector ¬eld X itself

will be called a complete vector ¬eld. In this case FlX is also sometimes called

t

exp tX; it is a di¬eomorphism of M .

The support supp(X) of a vector ¬eld X is the closure of the set {x ∈ M :

X(x) = 0}.

Lemma. Every vector ¬eld with compact support on M is complete.

Proof. Let K = supp(X) be compact. Then the compact set 0 — K has positive

distance to the disjoint closed set (R—M )\D(X) (if it is not empty), so [’µ, µ]—

K ‚ D(X) for some µ > 0. If x ∈ K then X(x) = 0, so FlX (t, x) = x for all t

/

and R — {x} ‚ D(X). So we have [’µ, µ] — M ‚ D(X). Since FlX (t + µ, x) =

FlX (t, FlX (µ, x)) exists for |t| ¤ µ by theorem 3.7, we have [’2µ, 2µ]—M ‚ D(X)

and by repeating this argument we get R — M = D(X).

So on a compact manifold M each vector ¬eld is complete. If M is not

compact and of dimension ≥ 2, then in general the set of complete vector ¬elds

on M is neither a vector space nor is it closed under the Lie bracket, as the

2

following example on R2 shows: X = y ‚x and Y = x ‚y are complete, but

‚ ‚

2

neither X + Y nor [X, Y ] is complete.

3.9. f -related vector ¬elds. If f : M ’ M is a di¬eomorphism, then for any

vector ¬eld X ∈ X(M ) the mapping T f ’1 —¦ X —¦ f is also a vector ¬eld, which

we will denote f — X. Analogously we put f— X := T f —¦ X —¦ f ’1 = (f ’1 )— X.

But if f : M ’ N is a smooth mapping and Y ∈ X(N ) is a vector ¬eld there

may or may not exist a vector ¬eld X ∈ X(M ) such that the following diagram

commutes:

wu

u

Tf

TM TN

(1) X Y

w N.

f

M

De¬nition. Let f : M ’ N be a smooth mapping. Two vector ¬elds X ∈

X(M ) and Y ∈ X(N ) are called f -related, if T f —¦ X = Y —¦ f holds, i.e. if diagram

(1) commutes.

Example. If X ∈ X(M ) and Y ∈ X(N ) and X — Y ∈ X(M — N ) is given by

(X — Y )(x, y) = (X(x), Y (y)), then we have: