D± f

±

D± N

DM

(1) X Y

w N.

f

M

32.9

32.12 32. Vector ¬elds 329

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

C ∞ (M ← D± M ) and Y ∈ C ∞ (N ← D± N ) are called f -related, if D± f —¦ X = Y —¦ f

holds, i.e. if diagram (1) commutes.

32.10. Lemma. Let Xi ∈ C ∞ (M ← DM ) and Yi ∈ C ∞ (N ← DN ) be vector

¬elds for i = 1, 2, and let f : M ’ N be smooth. If Xi and Yi are f -related for

i = 1, 2, then also »1 X1 + »2 X2 and »1 Y1 + »2 Y2 are f -related, and also [X1 , X2 ]

and [Y1 , Y2 ] are f -related.

Proof. The ¬rst assertion is immediate. To prove the second we choose h ∈

C ∞ (N, R), and we view each vector ¬eld as operational. Then by assumption

we have Df —¦ Xi = Yi —¦ f , thus:

(Xi (h —¦ f ))(x) = Xi (x)(h —¦ f ) = (Dx f.Xi (x))(h) =

= (Df —¦ Xi )(x)(h) = (Yi —¦ f )(x)(h) = Yi (f (x))(h) = (Yi (h))(f (x)),

so Xi (h —¦ f ) = (Yi (h)) —¦ f , and we may continue:

[X1 , X2 ](h —¦ f ) = X1 (X2 (h —¦ f )) ’ X2 (X1 (h —¦ f )) =

= X1 (Y2 (h) —¦ f ) ’ X2 (Y1 (h) —¦ f ) =

= Y1 (Y2 (h)) —¦ f ’ Y2 (Y1 (h)) —¦ f = [Y1 , Y2 ](h) —¦ f.

But this means Df —¦ [X1 , X2 ] = [Y1 , Y2 ] —¦ f .

32.11. Corollary. Let D± be one of the following functors D, D(k) , T . Let

f : M ’ N be a local di¬eomorphism so that (Tx f )’1 makes sense for each x ∈ M .

Then for Y ∈ C ∞ (N ← D± N ) a vector ¬eld f — Y ∈ C ∞ (M ← D± M ) is de¬ned

by (f — Y )(x) = (Tx f )’1 .Y (f (x)), and the linear mapping f — : C ∞ (N ← Dβ N ) ’

C ∞ (M ← Dβ M ) is a Lie algebra homomorphism, i.e. f — [Y1 , Y2 ] = [f — Y1 , f — Y2 ],

where Dβ is one of D, T , D[1,∞) .

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

de¬ned on an interval J. It will be called an integral curve or ¬‚ow line of a kinematic

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

For a given kinematic vector ¬eld integral curves need not exist locally, and if they

exist they need not be unique for a given initial value. This is due to the fact that

the classical results on existence and uniqueness of solutions of equations like the

inverse function theorem, the implicit function theorem, and the Picard-Lindel¨f o

theorem on ordinary di¬erential equations can be deduced essentially from one

another, and all depend on Banach™s ¬xed point theorem. Beyond Banach spaces

these proofs do not work any more, since the reduction does no longer lead to a

contraction on a metrizable space. We are now going to give examples, which show

that almost everything that might fail indeed fails.

32.12

330 Chapter VII. Calculus on in¬nite dimensional manifolds 32.12

Example 1. Let E := s be the Fr´chet space of rapidly decreasing sequences.

e

Note that s = C ∞ (S 1 , R) by the theory of Fourier series. Consider the continuous

linear operator T : E ’ E given by T (x0 , x1 , x2 , . . . ) := (0, 12 x1 , 22 x2 , 32 x3 , . . . ).

The ordinary linear di¬erential equation x (t) = T (x(t)) with constant coe¬cients

and initial value x(0) := (1, 0, 0, . . . ) has no solution, since the coordinates would

have to satisfy the recursive relation xn (t) = n2 xn’1 (t) with x1 (t) = 0, and hence

we must have xn (t) = n!tn . But the so de¬ned curve t ’ x(t) has only for t = 0

values in E. Thus, no local solution exists. By recursion one sees that the solution

for an arbitrary initial value x(0) should be given by

n

tn’i

n! 2

xn (t) = xi (0) .

i! (n ’ i)!

i=0

If the initial value is a ¬nite sequence, say xn (0) = 0 for n > N and xN (0) = 0,

then

N

tn’i

n! 2

xn (t) = xi (0)

i! (n ’ i)!

i=0

N

(n!)2 12

xi (0) (n’N )! tN ’i

tn’N

= i! (n’i)!

(n ’ N )! i=0

N ’1

(n!)2 12 12

|xi (0)| (n’N )! |t|N ’i

|t|n’N

|xn (t)| ≥ |xN (0)| ’

N! i! (n’i)!

(n ’ N )! i=0

N ’1

(n!)2 2 12

|xi (0)||t|N ’i

|t|n’N 1

≥ |xN (0)| ’ ,

N! i!

(n ’ N )! i=0

where the ¬rst factor does not lie in the space s of rapidly decreasing sequences,

and where the second factor is larger than µ > 0 for t small enough. So at least for

a dense set of initial values this di¬erential equation has no local solution.

This also shows that the theorem of Frobenius is wrong in the following sense: The

vector ¬eld x ’ T (x) generates a 1-dimensional subbundle E of the tangent bundle

on the open subset s \ {0}. It is involutive since it is 1-dimensional. But through

points representing ¬nite sequences there exist no local integral submanifolds (M

with T M = E|M ). Namely, if c were a smooth non-constant curve with c (t) =

f (t).T (c(t)) for some smooth function f , then x(t) := c(h(t)) would satisfy x (t) =

T (x(t)), where h is a solution of h (t) = 1/f (h(t)).

Example 2. Next consider E := RN and the continuous linear operator T : E ’ E

given by T (x0 , x1 , . . . ) := (x1 , x2 , . . . ). The corresponding di¬erential equation

has solutions for every initial value x(0), since the coordinates must satisfy the

recursive relation xk+1 (t) = xk (t), and hence any smooth function x0 : R ’ R

(k) (k)

gives rise to a solution x(t) := (x0 (t))k with initial value x(0) = (x0 (0))k . So

by Borel™s theorem there exist solutions to this equation for all initial values and

the di¬erence of any two functions with same initial value is an arbitrary in¬nite

32.12

32.13 32. Vector ¬elds 331

¬‚at function. Thus, the solutions are far from being unique. Note that RN is

a topological direct summand in C ∞ (R, R) via the projection f ’ (f (n))n , and

hence the same situation occurs in C ∞ (R, R).

Note that it is not possible to choose the solution depending smoothly on the initial

value: suppose that x is a local smooth mapping R—E ⊃ I—U ’ E with x(0, y) = y

and ‚t x(t, y) = T (x(t, y)), where I is an open interval containing 0 and U is open

in E. Then x0 : I — U ’ R induces a smooth local mapping x0 ∨ : U ’ C ∞ (I, R),

which is a right inverse to the linear in¬nite jet mapping j0 : C ∞ (I, R) ’ RN = E.

∞

Then the derivative of x0 ∨ at any point in U would be a continuous linear right

∞

inverse to j0 , which does not exist (since RN does not admit a continuous norm,

whereas C ∞ (I, R) does for compact I, see also [Tougeron, 1972, IV.3.9]).

Also in this example the theorem of Frobenius is wrong, now in the following

sense: On the complement of T ’1 (0) = R — 0 we consider again the 1-dimensional

subbundle generated by the vector ¬eld T . For every smooth function f ∈ C ∞ (R, R)

∞

the in¬nite jet t ’ jt (f ) is an integral curve of T . We show that integral curves

through a ¬xed point sweep out arbitrarily high dimensional submanifolds of RN :

Let • : R ’ [0, 1] be smooth, •(t) = 0 near t = 0, and •(t) = 1 near t = 1. For

each (s2 , . . . , sN ) we get an integral curve

s2 s3 sN

•(t)(t ’ 1)2 + •(t)(t ’ 1)3 + · · · + •(t)(t ’ 1)N

t ’ jt t +

2! 3! N!

connecting (0, 1, 0, . . . ) with (1, 1, s2 , s3 , . . . , sN , 0, . . . ), and for small s this integral

curve lies in RN \ 0.

Problem: Can any two points be joined by an integral curve in RN \ 0: One has to

¬nd a smooth function on [0, 1] with prescribed jets at 0 and 1 which is nowhere

¬‚at in between.

Example 3. Let now E := C ∞ (R, R), and consider the continuous linear operator

T : E ’ E given by T (x) := x . Let x : R ’ C ∞ (R, R) be a solution of the

‚ ‚

equation x (t) = T (x(t)). In terms of x : R2 ’ R this says ‚t x(t, s) = ‚s x(t, s).

ˆ ˆ ˆ

Hence, r ’ x(t ’ r, s + r) has vanishing derivative everywhere, and so this function

ˆ

is constant, and in particular x(t)(s) = x(t, s) = x(0, s + t) = x(0)(s + t). Thus, we

ˆ ˆ

have a smooth solution x uniquely determined by the initial value x(0) ∈ C ∞ (R, R),

which even describes a ¬‚ow for the vector ¬eld T in the sense of (32.13) below. In

general however, this solution is not real-analytic, since for any x(0) ∈ C ∞ (R, R)

which is not real-analytic in a neighborhood of a point s the composite evs —¦x =

x(s+ ) is not real-analytic around 0.

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

FlX

A local ¬‚ow FlX for X is a smooth mapping M — R ⊃ U ’ ’ M de¬ned on a

’

c∞ -open neighborhood U of M — 0 such that

(1) U © ({x} — R) is a connected open interval.

(2) If FlX (x) exists then FlX (x) exists if and only if FlX (FlX (x)) exists, and

s t+s t s

we have equality.

32.13

332 Chapter VII. Calculus on in¬nite dimensional manifolds 32.15

(3) FlX (x) = x for all x ∈ M .

0

(4) dt FlX (x) = X(FlX (x)).

d

t t

In formulas similar to (4) we will often omit the point x for sake of brevity, without

signalizing some di¬erentiation in a space of mappings. The latter will be done

whenever possible in section (42).

32.14. Lemma. Let X ∈ X(M ) be a kinematic vector ¬eld which admits a local

¬‚ow FlX . Then each for each integral curve c of X we have c(t) = FlX (c(0)), thus

t t

X

there exists a unique maximal ¬‚ow. Furthermore, X is Flt -related to itself, i.e.,

T (FlX ) —¦ X = X —¦ FlX .

t t

Proof. We compute

FlX (’t, c(t)) = ’ ds |s=’t FlX (s, c(t)) + FlX (’t, c(s))

d d d

ds |s=t

dt

= ’ ds |s=0 FlX FlX (s, c(t)) + T (FlX ).c (t)

d

’t ’t

= ’T (FlX ).X(c(t)) + T (FlX ).X(c(t)) = 0.

’t ’t

Thus, FlX (c(t)) = c(0) is constant, so c(t) = FlX (c(0)). For the second assertion

’t t

we have X —¦ Flt = dt Flt = ds |0 Flt+s = ds |0 (FlX —¦ FlX ) = T (FlX ) —¦ ds |0 FlX =

X X X

d d d d

t s t s

X

T (Flt ) —¦ X, where we omit the point x ∈ M for the sake of brevity.

32.15. The Lie derivative. For a vector ¬eld X ∈ X(M ) which has a local

¬‚ow FlX and f ∈ C ∞ (M, R) we have dt (FlX )— f = dt f —¦ FlX = df —¦ X —¦ FlX =

d d

t t t t

X X—

X(f ) —¦ Flt = (Flt ) X(f ).

We will meet situations (in (37.19), e.g.) where we do not know that the ¬‚ow of X

exists but where we will be able to produce the following assumption: Suppose that

• : R — M ⊃ U ’ M is a smooth mapping such that (t, x) ’ (t, •(t, x) = •t (x))

is a di¬eomorphism U ’ V , where U and V are open neighborhoods of {0} — M

‚

in R — M , and such that •0 = IdM and ‚t 0 •t = X ∈ X(M ). Then we have

•’1 = ’X, and still we get dt |0 (•t )— f = dt |0 (f —¦ •t ) = df —¦ X = X(f ) and

‚ d d

t

‚t 0

similarly ‚t 0 (•’1 )— f = ’X(f ).

‚

t

Lemma. In this situation we have for Y ∈ C ∞ (M ← DM ):

|0 (•t )— Y

d

= [X, Y ],

dt

|0 (FlX )— Y

d

= [X, Y ],

t

dt

X—

= (FlX )— [X, Y ].

d

dt (Flt ) Y t

Proof. Let f ∈ C ∞ (M, R) be a function, and let ±(t, s) := Y (•(t, x))(f —¦ •’1 ),

s

which is locally de¬ned near 0. It satis¬es

±(t, 0) = Y (•(t, x))(f ),

±(0, s) = Y (x)(f —¦ •’1 ),

s

‚ ‚ ‚

‚t ±(0, 0) = Y (•(t, x))(f ) = ‚t 0 (Y f )(•(t, x)) = X(x)(Y f ),

‚t 0

|0 Y (x)(f —¦ •’1 ) = Y (x) ‚s |0 (f —¦ •’1 ) = ’Y (x)(Xf ).

‚ ‚ ‚

‚s ±(0, 0) = s s

‚s

32.15

32.17 32. Vector ¬elds 333

‚

‚u |0 ±(u, u)

Hence, = [X, Y ]x (f ). But on the other hand we have

’1

‚ ‚

‚u |0 ±(u, u) ‚u |0 Y (•(u, x))(f —¦ •u ) =

=

’1

‚

‚u |0 D(•u ) —¦ Y —¦ •u x (f )

=

( ‚u |0 (•u )— Y )x (f ),

‚

=

so the ¬rst two assertions follow. For the third claim we compute as follows:

X—

D(FlX ) —¦ D(FlX ) —¦ Y —¦ FlX —¦ FlX

‚ ‚

‚s |0

‚t (Flt ) Y = ’t ’s s t

= D(FlX ) —¦ D(FlX ) —¦ Y —¦ FlX —¦ FlX

‚

‚s |0

’t ’s s t

= D(FlX ) —¦ [X, Y ] —¦ FlX = (FlX )— [X, Y ].

’t t t

32.16. Lemma. Let X ∈ X(M ) and Y ∈ X(N ) be f -related vector ¬elds for a

smooth mapping f : M ’ N which have local ¬‚ows FlX and FlY . Then we have

f —¦ FlX = FlY —¦f , whenever both sides are de¬ned.

t t

—

Moreover, if f is a di¬eomorphism we have Flf Y = f ’1 —¦ FlY —¦f in the following

t t

sense: If one side exists then also the other and they are equal.

For f = IdM this again implies that if there exists a ¬‚ow then there exists a unique

maximal ¬‚ow FlX .

t

Proof. We have Y —¦ f = T f —¦ X, and thus for small t we get, using (32.13.1),

Y

—¦f —¦ FlX ) = Y —¦ FlY —¦f —¦ FlX ’T (FlY ) —¦ T f —¦ X —¦ FlX

d

dt (Flt ’t ’t ’t

t t

= T (FlY ) —¦ Y —¦ f —¦ FlX ’T (FlY ) —¦ T f —¦ X —¦ FlX = 0.

’t ’t

t t

So (FlY —¦f —¦FlX )(x) = f (x) or f (FlX (x)) = FlY (f (x)) for small t. By the ¬‚ow prop-

’t

t t t

erties (32.13.2), we get the result by a connectedness argument as follows: In the

common interval of de¬nition we consider the closed subset Jx := {t : f (FlX (x)) =

t

Y X

Flt (f (x))}. This set is open since for t ∈ Jx and small |s| we have f (Flt+s (x)) =

f (FlX (FlX (x))) = FlY (f (FlX (x))) = FlY (FlY (f (x))) = FlY (f (x)).

s t s t s t t+s

The existence of the unique maximal ¬‚ow now follows since two local ¬‚ows have to

agree on their common domain of de¬nition.

32.17. Corollary. Take X ∈ X(M ) be a vector ¬eld with local ¬‚ow, and let

Y ∈ C ∞ (M ← DM ). Then the following assertions are equivalent

(1) [X, Y ] = 0.

(2) (FlX )— Y = Y , wherever de¬ned.

t

If also Y is kinematic and has a local ¬‚ow then these are also equivalent to

(3) FlX —¦ FlY = FlY —¦ FlX , wherever de¬ned.

t s s t

Proof. (1) ” (2) is immediate from lemma (32.15). To see (2) ” (3) we note

X—

that FlX —¦ FlY = FlY —¦ FlX if and only if FlY = FlX —¦ FlY —¦ FlX = Fl(Flt ) Y by

’t

t s s t s s t s

X—

lemma (32.16), and this in turn is equivalent to Y = (Flt ) Y , by the uniqueness

of ¬‚ows.

32.17

334 Chapter VII. Calculus on in¬nite dimensional manifolds 32.18

32.18. Theorem. [Mauhart, Michor, 1992] Let M be a manifold, and let •i :

R — M ⊃ U•i ’ M be smooth mappings for i = 1, . . . , k such that (t, x) ’

(t, •i (t, x) = •i (x)) is a di¬eomorphism U•i ’ V•i . Here the U•i and V•i are open

t

‚

neighborhoods of {0} — M in R — M such that •i = IdM and ‚t 0 •i = Xi ∈ X(M ).

t

0

j j ’1 j

i ’1

i j i i

We put [• , • ]t = [•t , •t ] := (•t ) —¦ (•t ) —¦ •t —¦ •t . Then for each formal bracket

expression P of length k we have

|0 P (•1 , . . . , •k )

‚

for 1 ¤ < k,

0= t t

‚t

1 ‚k 1 k

k! ‚tk |0 P (•t , . . . , •t ) ∈ X(M )

P (X1 , . . . , Xk ) =

as ¬rst non-vanishing derivative in the sense explained in step (2 of the proof. In

particular, we have for vector ¬elds X, Y ∈ X(M ) admitting local ¬‚ows

(FlY —¦ FlX —¦ FlY —¦ FlX ),

‚

0= ’t ’t t t

‚t 0

1 ‚2 Y X Y X

2 ‚t2 |0 (Fl’t —¦ Fl’t —¦ Flt —¦ Flt ).

[X, Y ] =

Proof. Step 1. Let c : R ’ M be a smooth curve. If c(0) = x ∈ M , c (0) =