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____~

N

ON THE SPACE OF RIEMANNIAN METRICS by David G. Ebin

A.B.,

Harvard University

(June 1964) Submitted in Partial Fulfillment of the Requirements for the Degree of DOCTOR OF PHILOSOPHY at the MASSACHUSETTS INSTITUTE OF TECHNOLOGY September, 1967

Signature of Author............................... Department of Mathematics, August 21,

Certified by

/4".

Accepted by......

F... .,vvvvyL........-o

1967

Thesis Supervisor

Thesis Supervisor

...

****a* ......

*

Chairman, Departmental Committee on Graduate Students

__

ZC-- - --~----r-r--·----

-ii-

On the Space of Riemannian Metrics by David G. Ebin

Submitted to the Department of Mathematics, in partial fulfillment of the requirements for the degree of Doctor of Philosophy.

ABSTRACT

The space of Riemannian metrics of a smooth compact manifold is investigated, particularly the natural action of the group of diffeomorphisms on that space. Both the space and the group are enlarged slightly, so that they can be endowed with differentiable structures. The orbits of the group are studied and a "slice" in the space of metrics transversal to these orbits is constructed. Also, the isotropy groups of the action are considered --they are trivial for an open dense subset of the space of metrics.

Thesis Supervisor: Title:

Isadore M. Singer Professor of Mathematics

-iii-

ACKNOWLEDGEMENTS

The author wishes to thank Professor I. M. Singer for his guidance, encouragement, and highly fruitful ideas.

Also, he is most grateful to Mrs. Mary Perron who

did the typing, and Miss Harriet Fell who magnanimously donated her time to fill in the handwritten portions of this work.

;"11

--

-1-

ON THE SPACE OF RIEMANNIAN METRICS

I.

Introduction and Outline of Arguments. Let

a9

be a compact differentiable manifold,

M

M , and ht the set of

the group of diffeomorphisms of smooth Riemannian metrics on

4

M .

acts on

XJ.from

the right in a natural way; i.e., there exists a natural map

A:h

X A ->

A (TIA(ty)

=

0- such that lr jt

A(rijy)

,

'y Es

It is the purpose of the present work to investigate this action and to discuss its properties. 'yE

we propose to construct at each point

which is transversal to the orbit of @& (definition of slice below). subgroup IV =

(Tj

I e

of s8

y

a slice

through

in

hl

A(q,y) = y)

sufficiently near to

.

y

defines a

called the isotopy group at

is trivial for some fixed V'

Each

In particular,

y ---

We shall also show that if V, V in

I

'

I

is also trivial for all

Z.

We first introduce some notation:* Smooth means differentiable arbitrarily many times. M is a smooth manifold,

T(M)

its tangent bundle,

We use primarily the notation of 12.

T*(M)

the

-2-

(T*)k

cotangent bundle, k-tensors, and

the subbundle of symmetric covariant M

is the tangent space to

Tp(M)

If p s M

tensors. at

SkT*

the bundle of covariant

If

p ; similar notation for cotangent bundle, etc. f: M -> N

is also a manifold and

N

is smooth, Tf: T(M) -> T(N)

is the induced map on the tangent bundle and Tpf: Tp(M) -> Tf(p)(N)


is defined by restricting

is a Riemannian metric, and Tp(M) .

product on p

bundle,

follows:

the equation

y s

A:

0 X• -> A

t , ?t an

p

is defined as

, then

A(n,,y)

A(0,)p (XY) = Y~(p)(TqX,

p a M , X,Y e Tp(M)

is the inner

y in the fibre over

is the Value of

If

>p

y is a section of some vector

If

The right action


M

a smooth map with

the right action property stated above. G(x) = A(G,x) , the orbit of

G

Fix

through

x s M ; let

x ; let

equal

(g e GIA(g,x) = x) , the isotropy group of

Since

A is continuous,

is a Lie subgroup of

G .

Gx If

G a

is closed in r: G -> G/Gx

G , so is the

Gx x

Gx

-3-

projection map, there is a unique topology (the quotient topology) on

G/Gx

With this topology such that

G/Gx

f: G/Gx ->

7: 0 ->

f

such that

,

A

coset

0 in

~C (U))'tX

(7r

A

R U

is a neighborhood of the trivial

is the identity map on

induces a smooth injection

space of

G/Gx , so

T4

M

G , and

(see 5, p. 110).

G/Gx -> M

4(G/G ) is a submanifold of

defined

X , and

is a diffeomorphism, whose range

(See 4, pp. 105, 108)

notation, for any

M .

Using the above

x E M , there exists a submanifold

S

such that:

1) S 2)

G/Gx , and

F

is invariant under action of

A(g,S) n S

3) If

then

4:

U

is compact and therefore closed in

Slice Theorem:

of

: U -> G

is injective on every tangent

G/Gx -> 4(G/Gx) = G(x)

G(x)

Also there exists a local cross

there is a smooth map ;

4(gGx) = A(g,x) .

by

is smooth if and only if

Xj (0) = Id , the identity element of

such that

4:

G/Gx

is continuous and open.

has a natural manifold structure

is smooth.

section; that is, if

7

0 0

implies

X( : U -> G F: U X S -> M

Gx

g s Gx

is a local cross section for is defined by

F(u,s) = A() (u),s) ,

is a diffeomorphism onto an open set in

M .

__ ____~_~~_~ _~_____

-4-

Outline of Proof: g: M ->

a map

For

g E G , we will consider

M defined by

g(x) = A(g,x). g -i

map with inverse the map defined by Tg: T(M) -> T(M) Let

G

< , > .

Let

Exp

G ; i.e., g: M -> M

be the exponential map of

corresponding to the Riemannian structure. such that

G(x)

Exp(V)

is defined,

for all

V

x

in

N

and

E-neighborhood of

Exp G(x)

S = N n Tx(M) ; that is

Tx(G(x))

r

N in S

M

so the Riemannian v(G(x)) over

E > 0

Tx(M)

of

a slice (see 4, p. 108).

Also, if

M

by

, Exp(V)

such

is defined

(see 12, pp. 73-75).

is the set of vectors at

S

F: G X S ->

is

is a diffeomorphism onto

It is easy to check that

that:

Exp(Tg(V))

E , which are perpendicular to

with length less than

the subspace

Then if

is compact, one can find an

N = (V E v(G(x))I < E2)

Let

Amon

G(x)

M

T(M) .

that if

an

M

M defines a normal bundle

G(x) , which is a subset of Since

becomes

g Exp(V) = Exp(Tg(V)) .

is a closed submanifold of

structure on

By

we can find a metric that is

invariant under the action of

also defined, and

is a smooth

be the induced map on the tangent bundle.

integrating it over

V E T(M)

as

g-l . Let

M have a Riemannian metric,

an isometry.

g

, and let

S = Exp S

has the three properties of Gx = [Id) , it follows

F(g,s) = A(g,s)

is a diffeomorphism

_

iC~ 1__·__

·1 -__--_

111~---1 ~.~___

1.1_~_.

-5-

onto a neighborhood of this neighborhood,

G

G(x)

in

M , and for any

y

in

= (Id)

When we try to generalize this theorem to the action of the group or

of diffeomorphisms of a manifold

the space of metrics ficulties.

L

of

M , we encounter several dif-

Our foremost problem is that 4

(nor is it locally compact).

M , on

is not compact,

Therefore, we cannot integrate

with respect to Haar measure to construct a metric on which is preserved under the action of not know that an orbit o8 (x) of a cannot immediately conclude that or that the map

4: 4 / 8x -> B(x)

6

77L

. Also we do

is compact, so we (x) W is closed in '

,

is a homeomorphism.

Furthermore there are technical difficulties regarding the topologies on the spaces

0 and ?i

.

is the space of smooth symmetric covariant M , and

Ce(S 2T*)

Ce(S 2 T*)

If

2-tensors on

has the topology of uniform convergence

in all derivatives, then it is a Frechet space.

If we

give 7h. this same topology, it is an open convex cone in eC(S

2

T*) --- an open set of a Frechet space.

If we give ·&

the topology of uniform convergence in all derivatives, we find that 6

is locally like a neighborhood in

Ce(T) ,

the space of smooth vector fields, and this is also a Frechet space (see 5 or 6).

__·_~

-6-

The usual proof of the slice theoxem uses the fact that the manifold

M

has an exponential map,

Exp , which is

a diffeomorphism near the zero section of the tangent bundle. However, for Frechet spaces, the implicit function theorem and the local existence theorem for ordinary differential equations are not true in general (see 12, p. vi).

There-

fore we do not know that there is an exponential map with the above property, so we cannot use the usual proof with these spaces. Because of these technical difficulties, we enlarge the sets 7

and

somewhat, so that they are infinite

dimensional manifolds, loelly like meighborhoods in Hilbert spaces. These enlarged spaces, which we call Hs

are spaces of

maps ---

48

s

that each partial is square integrable. s

or "

s

s

maps which have partial der-

ivatives defined almost everywhere up to order

to Z

and "

s

such

A tangent space

at a*l point looks like a Sobolev space,

HS(E) ; i.e., a space of sections of some vector bundle on

E

M , such that each section has square integrable partial

derivatives up to order

s

(see 16).

Also we can construct in a natural way, a Riemannian metric on " The action

s A

which is invariant under the action of extends to a continuous map

.

A: O s+1 x•s->

-7-

y E9fls

At any point by

7T(T

) = A(r,y) .

q E

Also, if for T (

s+l)

amd

s+1 _>

If

s+l

y

T1 P7

ential operator

a .

,

s)

lt is smooth.

with the appropriate

becomes a first order linear differIt turns out that

symbols so its range is closed in

The isotropy group

£0 s+l

a

induced manifold structure. ->1 2S

.

has injective

T?P ()(.

s)

(also called

compact Lie group, and the factor space b

$:0 s +

is defined

we identify the tangent spaces

T*()(T

Sobolev spaces,

is in 9

s

*7

I )

s+1/Iy

is a has an

induces a map

$4 is an injective immersion, but

we do not know that it is a homeomorphism onto a closed

orbit (as in the classical case). It does, however, induce a normal bundle v on Ss+ 1 /I7 , and the exponential, exp on 1) s coming from the Riemannian structure of section of

9j

s

is defined near the zero

v , though it is not necessarily a diffeomor-

phism. Also, we know that any fixed smooth map

T.*:01es

is an isometry. V

in

T(ai

s )

->V7s

by

ZS s+1

defines a

q*(y) = A(rT,y) , and

q*

Therefore, if exp is defined on a vector exp is defined on

Ti*(V) , and

q*exp(V) = exp Tn*(V) Combining the above information, we get the following restricted version of the slice theorem.

-8-

If A:65 s+l X Vs _>9sn

Theorem:

above, for each

is the action defined

, there is a submanifold

'yea

S

of

a such that: sl T EI

1) If 2)

There is a neighborhood

such that if 3) T:

A(q,S) = S

If

s+l _

T

s V and

A(i,S) n S

C( : U ->D s+l , and

s+1/I

F(u,s) = A(3 (u),s) , then neighborhood of

V of

I

inB s+1

/ 0 then

T E I.y

is a local cross section for F: U X S ->7s F

is defined by

is a homeomorphism onto a

.

In Section II, we develop some tools of calculus and discuss spaces of

Hs

functions on open sets in Euclidean

space. In Section III, we construct the infinite dimensional manifold of manifold.

Hs

maps from a compact manifold to some other

We define the group Z

s

and discuss its

properties. Section IV is about the manifold

s .

Using differ-

ential operators we define a Riemannian metric

p

s .

on

In Section V we discuss the group action A: B s+l x1

S _>

js , check its smoothness, and define

a manifold structure on 4 s+1/I In Section VI we show that inJective immersion.

1a

:

S+/I -> s+1

s

is an

-9-

In Section VII we show that our Riemannian metric is invariant under the action of 4 s+l , and look at the normal bundle on O

s+1/I

induced by the immersion

.

In Section VIII we state and prove the slice theorem itself. In Section IX we discuss the smooth situation --- we show that

$.: Z/I

closed subset of in %

-> 27

is a homeomorphism onto a

, and we prove that the set of metrics

with trivial isometry group is open , and dense in 31. In Section X we make a few suggestions as to what

further research might be done, extending the present work.

-----------mý

-10-

Hs Functions.

II.

The purpose of this section is to define a set of

kIn to IR m

functions from

,

Hs

called

functions,

(s a positive integer), and to put a topology on this set.

Later we will generalize the definition to a set of

maps from one manifold to another and construct two examples of such sets called Z s Let struct give

U

HS(U)

to

.

be a bounded open set in a n . We shall conHs

HS(U) , the

functions from

U to

1Rm and

the structure of a Hilbert space.

Ce(U)

Let

and *s

be the set of smooth functions from

U

nRmwhich are bounded and have all derivatives

bounded, and let

CO(U)

be the subset consisting of those

functions whose support is contained in

U .

These are

linear spaces under the operations of pointwise addition and scalar multiplication.

product on

d~(U)

Definition 1:

If

(l,a2,...,an)

,

I

(I

a 1

We shall define an inner

a)n 2

functions on gtn

f e and

(U) , n Ial = 7

a

an

n-tuple of integers

i '

i=1

a n

,

h

di

t ILCL qu

-11-

Daf e C' (U) .

Remark:

Definition 2:

C'(U)

Let

be a bilinear function on

=

f

defined by: (fg)

where

ls

U

f

Da g

dx

defines a norm on

Definition 3:

HS(U)

respect to the norm in

HS(U)

in the norm

n .

It

is

( )s is positive definite, so

= (f,f)/2

dmo(U)

sal s

D

refers to the usual measure on

dx

clear that

IIf

< , >s

II

is the completion of

II ItI and

HS(U)

(see 2, p. 192).

II5

Ce(U) C"(U)

with

is the closure of

Both are Hilbert spaces

We shall write

HS(U,

R

m)

when we

We show an elementary property of the space

HS(u) . 0

0

wish to explicitly specify the range.

Proposition 4: Uc V .

i:

Let

U

V

be bounded open sets of

' n

Then there is a natural inclusion

Hs (U) -> Hs (V) defined by 0

0

i(f)(v)

=

0 v vU

f(v)

v

U

which is a continuous linear map. Proof: map. the

~

i:

CO(U) -> CO(V) defined as above is a linear 0 0 It is clearly bounded (in fact norm preserving) in

Hs

norms on

U

and

V . Therefore it extends to

-12-

0

f: 9 ->

0

9

Now let

be any open set in ft n

and let

m

Definition 5: p

q.e.d.

i: HS(U) -> HO(V).

the map

iff there are bounded neighborhoods

V C U , and a

C

function p

support (p) C U ,

f

is

Hs

at

U, V

of

p

with

p E 9 , we say that

Given

p: U ->

IR

1

identically

such that

on

V

pf s H0(U)

and

We now procede to demonstrate some elementary properties of differentiation which we will need in our construction of spaces of Let let

U

iff

functions on manifolds. E

be an open set in the Banach space

f: U -> F ,

Definition 6:

HS

F

and

a Banach space.

A function from R to

1k

is called

0(t)

O(t)/t = 0

lim

t -> 0 Definition 7:

Assuming

horizontal at

0

U

contains

iff for any neighborhood

F , there is a neighborhood f(tU') Definition 8:

0 , we say

f: U -> F

U'

c

of

0

in

f

V

of

0

U

such that

is differentiable* at

p e U

___ __

I

notion

is

often

in

O(t)V

and only if there exists a continuous linear map

This

is

called

the

Frechet

derivative.

if

A: E -> F

---------------0

-13-

f(p + x) = f(p) + A(x) + h(x)

such that

of

in

0

E

II All

=

F .

to

max

ixill

It is a Banach space under the norm

11Axl

2

(see 13, PP. 4-5).

Definition 9:

Df: U -> L(E,F)

is differentiable at

p s U

Ck

is

f

if and only if

exists and is continuous.

smooth if

f

is

Ck

Dkf: U -> L(E,L(E '''

Let from

Lk(E,F)

IIAl

=

max

Ixil=1

Ck -

is

Df

...

L(E,F))

is

f

Of course it

if and only if is

f

Also we say

.

and we

.

Ck

is

f

Df

C

or

Ck

(k - L's) .

)

be the set of continuous multi-linear maps

EX EX ... X E

Definition 10:

D2f = D(Df)

k . If

for all

If

Dkf = D(Dk-lf)

follows from this definition that Dkf

Df(p) = Dp f

C1 .

is

f

we define

Proceeding inductively we write say

U

is defined by

is continuous we say that

Df

Assume

is differentiable at every point of

f: U -> F

If

be the space of continuous linear maps

L(E,F)

Let

Dpf .

and is denoted

f

is called the

A

F . In this case

into

E

derivative of

from

0

is a horizontal function from some neighborhood

h

where

near

x

for

into

F

(k - E's).

We define a norm on

(A(x 1 ,x 2 #** Xk)

Lk(E,F)

where

xi

by

E B

for all

i

-14-

Proposition 11:

Lk(E,F)

is a Banach space under this

norm and there is a canonical norm-preserving isomorphism

Q: Lk(E,F) -> L(E,L( Proof:

(L(E,F)) ...

*..

If A a Lk(E,F)

(

)

let

*** ((Q(A)(xl))(x 2 )) .. " )(Xk) = A(x1 ,x2 , *o* Xk)

It is easy to see that

Q is a norm-preserving isomorphism

(see 13, p. 5). Remark:

Because of the above proposition we can identify

L(E,L(E,

***

L(E,F))

as a map from

U

... )

with

Lk(E,F)

and consider

Df

Lk(E,F)

to

Another concept of differentiability is the following: Definition 12:

f: U -> F

has Gateaux dertvative at

lim [f(f(u+tv) - f(u)] t -,0 We write this limit as df(u,v) in direction

Lemma 13:

If

derivatives at direction

f: U -> V ,

x

df(x,v)

g: V -> F

in direction

v

respectively.

equals

dg(f(x), df(x,v)) .

Proof:

Usual chain rule argument.

Lemma 14: u e U

1

v iff

v sE

and

df

exists in

have Gateaux

and at

f(x)

d(gof)(x,v)

If f has Gateaux Derivative

and all

u

in exists and

df(u,v)

for all

defines a continuous map

F.

-15-

4: on

U -> L(E,F) U

$(u)(v) = df(u,v) , then

See 19,

p.

C1

is

6.

dkf(x,vl ...

Let

f

$= Df .

and

Proof:

by

Vk)

be the iterated

kth

Gateaux

derivative. Corollary 15:

If

f

x E U ,

exists for all

Ck 1

is

and

(vi ) C E

dkf(x,v

vk)

such that the map

$k: U -> Lk(E,F)

defined by $k(n)(vl ... vk) = dkf(u,vl 1 k and D f =k() . is continuous, then f is C Proof: dtf(u,v1

Assume the corollary for ""* Vt) = d(Dt-lf(v 1

by the lemma is

bounded

Du(Dt-lf(vl ..

t-linear map in

and equals

dtf(u,v1

**.

vi vt)

k = t-l , then

..

vt-1l))(u,v k ) . vt _))(vt) 1

. Therefore The case

This

which is a

Dt f u

t = 1

exists

is shown

in the above lemma, so this proves the corollary. This lemma and corollary will be used to show smoothness of maps from one space of functions to another. This completes our discussion of the derivatives of a single map.

We go on to develop the rules for differ-

entiating a composition of mappings. Proposition 16: f: U -> V ,

_---

Let

g: V -> G

U, V

be open sets of

(E,F,G

E, F ;

Banach spaces).

Then if

v

P

i-

-

.h..n

Dp (g o f) = Df(p)g * Dpf

where

composition of linear maps Proof:

E bF

Definition 17:

Proposition 18: at a point

p

II e + f

and

Df(p)g: F -> G

If

f E ,

of

Proof:

and f4

g g

E

and

F

(a Banach

If = (I e 1I 2 + 11 f 112)/2). g: G -> F , then

is defined by

Dp(f O g) = D f

(f 0 g)(x) = f(x) + g(x) (above) are differentiable

is differentiable at

p ,

DDpg .

First we note that the sum of two horizontal maps

is horizontal.

Let

0

in

E

0

in

E

Since

Then given

h2 : G -> F

V

and

F

U2

be defined

a neighborhood of

F , there exists neighborhoods

h1 , h2

and

hl: G -> E ,

0 .

and horizontal at

2

indicates the

Dpf: E -> F

f: G -> E ,

If

f D g: G -> E 0F

h 2 (tU

"

*

be the direct sum of

space under the norm

U1

"

Standard (see 17, p. 17).

Let

and

r

-

respectively such that

V1

and

of

V1 4 V2 C V .

are horizontal there exist neighborhoods of

) _c O(t)V2

0

in

.

Let

G , such that U = U1 n U2 .

hl (tU 1 ) _c O(t)V 1 , Then

h1 ) h2(tU) C 0(t)(V1 0 V2) C 0(t)V .

hi

V2

9 h2 is horizontal.

Our proposition now follows from the computation:

-17-

g)(p + x) = f(p + x) + g(p + x)

(fr

=f(p) + Dp f(x) + hl(x) + g(p) + Dpg(x) + h 2 (x) = (f

g)(p) + (D f 0 Dpg)(x) + (hl

h 2)(x)

We now prove a slight generalization of the formula for the derivative of the product of two functions. be Banach spaces; let bilinear map. A, B

Ck

Let

g: E X F -> G

U, V

Let

E,F,G

be a continuous

be open sets in Banach spaces

respectively, and let

f: U -> E

h: V -> F

be

functions.

Proposition 19:

g(f,h): U X V -> G

g(f,h)(u,v) = g(f(u), h(v))

D(u,v)(g(fh))(a,b)

Ck , and

is

= g(Duf(a),

defined by

h(v)) + g(f(u), Dvh(b))

a s A , b s B , u s U , v s V

where

Proof:

This is analogous to the standard proof for the

derivative of a product (see 5a, p. 167-169). Let a map by composition,

h: V -> G

and

F

are

1

g(X,Y) ->

C2

where

X * Y ,

and assume

U

V

and

be defined f: U ->

are open sets of

V ,

E

respectively.

Corollary 20: Proof:

g: L(E,F) X L(F,G) -> L(E,G)

g

2 D.... u-(h o f) =

2

(Duf Df(u)

Duf) + Df(uh * D2 +h •fu Duf

is a continuous bilinear mapping, so use the

-18-

above proposition on

Du(h o f) = Df(u)h * Duf

composition rule to get Corollary 21:

and the

D(Df(u)h) = D2(u)h

Let the above

f

and

h

be

Ck . Then

Dk(h o f) is a sum of terms of the form: u DP(u)h

where

i+ 1 2 + ...

1

Proof:

*

i i2 (Du l f, D1u f,.*

i , Du

f)

ip = k .

We have the case

k = 2 . The proof for larger

k

is by a direct induction using the preceding Corollary and the two previous propositions. The following Corollary will be useful in studying Hs

the composition of Corollary 22: Drh

DS(h o f)

functions. is a sum of terms in

such that for each term,

equal to Proof:

t

or

r

Dtf

and

is less than or

[s/2]* + 1

This follows by inspecting the formula of the

preceding corollary. If we apply the arguments similar to these of the above three corollaries to the situation g: E X F -> G ,

[a]

I

g

a bilinear map.

f: U -> E ,

h: V -> F ,

We get:

means the greatest integer less than or equal to

a

-19-

i DS g(f,h)=

Proposition 23:

(s)* g(Dif, DS-ih) i=0

Proof:

D g(Dif, DJh) = g(Di+lf, DJh) + g(Dif, Dj+lh)

by the previous proposition.

The formula now follows from

the binomial theorem. Corollary 23a: g(Drf, Dth)

DS(g o (f,h))

where

is a sum of terms of form

t < [s/2] + 1 .

r or

If we now restrict our attention to the finite dimensional situation --- let Rn ,

U, V

be open in Rm

---, then

f: U-> V , h: V-> IP

DufsDf(u)h

and are

linear transformations represented by the matrix of first order partial derivative of

f , and

h

respectively, (see 58, pp. 170-171).

at

u

and

f(u)

More generally

Drf is a matrix of rth order partial derivatives of u Corollary 22 implies the following result: Proposition 24:

Du(h o f) is a matrix of polynomials in

the partial derivatives of than or equal to be

N Dh DAf

s .

where

f

and

h , of order less

Each term of such a polynomial will N

is an integer and

IaI < [s/2] + 1 or I

Euclidean space

U, V

U ,

be bounded open sets of

g e HS(V)

and

f

H(U)

.

0 Assume s large enough so that HCs/2](U) C Cl(U0 HEs/2](V) C Cl(V) , as inSoboley lemma. Also assume Dg isnon-singular at each point v of V. If (f)C: CO(U) (g n

c C(V),

and

gn -> g

c U) (gn(V) n in

HS(V)

such that

,

in

n

and in

C[s/2]+1(v)

HS(U) 0

then

H s (V)

fngn ->

fg

Proof:

We use proposition 24 combined with the Sobolev

lemma.

We know that since

and clearly

in

fngn -> fg

in

f

and 0 (U) CO

Therefore, we must only show that

1

fn -> f

g

are

CO ,

fg

is

CO

(see 1, pp. 7-8). fngn

converges in

HS(V)

-22-

T rikC4

chrr+h~~n

~bmrI-

the matrix

Dk (fngn )

~

1 I

t~ sr v~rr h

converges in the

L2

By proposition 24, any such entry is:

Ii

where lal < [s/2] + 1 or

sense on

V

N Daf n D gn

[s/2] + 1

'

Case I: I

Jat


Hs-[s/2]+1(U) C H[s/2]-l(U)

C CO(U)

Also,

sense.

D g n ->

f ID fnDg

Dg

in

L2

- Daf DgI 2 < I Daf +

f

in

Daf

by hypothesis. Therefore

Dggn - Daf D~gn

IDaf DPgn - Daf Dgl

+

fV ID gn

2

fV iDgn

- DBgl

Hence, if

1

ial

D g n ->

D g

in

L 2 , and

goes to zero for the same reason. is bounded because

max (ID a f - D f u u n HU U

2)

2

max (IDafu 2 ) f lDgn - D 12

is bounded because

max (IDufI2) u EU

2

IDP n

2)

Da

< max (IDaf

12

Dafn -> D f

CO

goes to zero for the same reason.

< [s/2] + 1 , Daffn Dg D

n,cn

converges. veg

-23-

Case II:

I I0 Sra V tg

n

,_ (v)In v gn

-

a g(v)

V

f

D ()f

Dg

V g ) vn

-

2 v gn

2

- I1

fV Da(v)f fUIDal

2

2 Idet

DvgI

T-23a-

Similarly

fVID~n(vfn

-

'L

D()1

max (Idet D~gI'

i lD~fn fu

-

DafI2

vsV +-

and

(max

~ldet

fIjDg,(v~

-

-max fidet v~V

Dvgnl-l)

Dgcvf

I

D gJ-l])

f U jD~tf .2 1

(max(v~Idet D g ~l)l

-max

vs V

1 ]) [Idet D gV-

fIDloaI UU

2

i i a i.

-24-

The lemma follows.

so these terms goto zero.

Hs

From here on, when dealing with shall always assume that

is

s

functions, we

large enough so that

the above lemma holds. Remarks:

1)

because ben"c0aus

Ha(V)

tion of

"Hs

fn

T-TH(

->

fg

in

Therefore, since the definiTheefre, since the rdefini-

function" is a purely local one (we will prove

this soon) we have shown that f o g

implies 2) k

s ,

at

Hs

p

Hs

f, g

whenever

at

Dpg

and

g(p)

p

is non-singular.

The proof of the above lemma shows that for Dk(fng n )

Dk(fg)

converges to

and the same for

Dkfn L2> D kf

Since

fg E HS(V)

means that

HS(V)

is complete. is complete

L2

in the

sense.

g , we find that

the composition rule for derivatives holds almost everywhere for the function 3)

If

g: V -> U

f o g has the property that any neigh-

borhood of the boundary of g

contains the image under

of a neighborhood of the boundary of

f s H(U) , 4)

If

k

s ,

V , then

for all

gof e Hs(V)

fn -> f

hypothesis that all

U

Dkf

formly to zero on

Dg

in

Cs(U) , we do not need the

is non-singular because then, for is bounded and U .

Hence

Dkf - Dkfn

f ID (v)fl2


0 Dk f - Dkg(v)fl 2 -> fvl Vg(vD 0 5) If of

f

the

we can say that the derivatives

f E HO(U)

U , they being

are defined almost everywhere on L2

limit of the corresponding derivatives of (fn) C Ce (U

where

)

and

O

f -> f n-

in

We now prove another property of

fn

H s (U

)

Hs

functions which

.

0

is also a consequence of the Sobolev lemma and proposition 25. Lemma 29: G

and f

Let

be a bilinear map; E, F,

0

Let

e E H0(U,E)

Let

are Euclidean spaces.

HS(U,F) .

Then

g: E X F -> G

be the diagonal map.

A: U -> U X U

g o (e,f) o A s HS(U,G)

and the map

0

(eq) -> g o (e,f) o A

,

is a continuous bilinear map

B: Hs(U,E) X Hs(U,F) -> HS(U,G) 0 0 0 Proof:

where Claim: A

Let

en -> e

[e ) C Co(U) ,

in

HS(U) ,

fn -> f

0

([f

in

Hs(V), O 0

c Co(V)

o A) C C (U) . (g(en,f ) n-

0

is smooth, it is clear that

Since

g

is smooth and

g(en,fn) o A

is smooth,

so we need only show that support (g o (en,fn) o A) c U But if

K 1 = support (en ) ,

support (g o (en,fn) o A)

K 2 = support (fn) , then

C K1 n K2

.

Therefore

(g o (en,fn) o A) _CCO(U) n~f-a

ii

..._-

' I~·

·

-26-

Now we wish to show that g o (en,fn) o A -> g o (e,f) o A in DA: U -> L( .

Rn X IRn)

n

HO(U)

DuA = A:

by

D2 A = 0 , the bilinear map with image

nX

(0) .

and

Therefore, by

the composition rule for derivatives and proposition 19, Du(g o (e,f) o A) = g(Due,f) + g(e,Duf)

and more generally

DS(g o (e,f) o A) = DS(g o (e,f)) o A .

By proposition 25,

DS(g o (e,f)) have form

IN

is a matrix of polynomials whose terms

p(g) Dae D f

< [s/2] + 1 .

such that

=

Xn

a1-

U

p ((g) g )

D Dn

< [s/2] + 1 or

Therefore, to show

g(en, n) o A -> g(e,f) o A in n

taI

H~(U)

D n --

we need only show 2

e Df

-> 0

But

< Ip(g) 2 1f IDen Dfn - Dae Dfn212 U

+ Ip(g) 2 fIDCe U DPfn - Dae D f12 Therefore, if

ijal

[s/2] + 1 ,

[IDe - Dae 2 Xn < Ip(g)21max us U De - De

UD fn

ap

2IDf

2 [IDue + Ip(g) lmax

2)

usU

f

U

f

n

- D fl 2

From here it follows as in the composition lemma that

Xn

-> 0. Xn

(1,f,n)

is symmetric in the triplets .' .

Therefore,

n

Xn

ap

->

0

T

(a,e,en) the case

and

..........

-27-

1II < [s/2] + 1 also.

Hence

g(e,f) o A s HS(U) 0

Now we check that

B: HS(U) x Hs(U) -> HS(U) is a 0 0 0 bilinear continuous map. B is obviously bilinear, because g

is. To check continuity we must show that there is a

constant

K

such that

II

I1 B(e,f)

_
JR

N2f

U

be a bounded set in Euclidean space

an

H

Hs

We first show that

Fix p

U . We know by the definition of

Af

that there are neighborhoods E

R )

Co(U p ,

I,

Therefore, lemma 29. so

Xf

Hs

OP pPf2

at

Up

)

E

and

function. Hs

functions

Vp

of

p

1

and

so

p p HS(U ) . Also

POp p. Furthermore,

C support (X2)

support (Pf)

= support (;2f)

is compact.

p e U , pick

ppVf e HS(U ).

Up , Vp

D support (D) .

corresponding pi f

E

Up

and

C Hs(U) Hs(Ui) 0 -

proposition 4).

0

HS(U)

pp ,

pp

Since support (x)

Call these

Ui

(pi) ,

V

and

by 1

so

cU

and

we can choose a finite number of points V = U V

E

0P 2 pp ý Vp

= support (A2f)

above, so that

ppf

HO(U )

ppA ppf

support (kf)

Now for each

and

as is compact

such that

call the

pi

under the natural inclusion (see

Therefore

(

1

Pi

) Vf

E

Hs(U)

0

.

Now consider the function

.

It is defined on

1:P V

since, for all

Pi(v) = 1

1

Also

.

X is smooth and support (pd C U

since

HO(Up',

p X

P Hence

is

)

is an

pp ( VP

such that

Pph E C (Up

But

Then

Hs (U) 0

E

Proof:

p

.

E Ce (U,l)

function, and

v E V , there is some

pi

support (X) C V , so

-i zPi

such that E

C(V)

0

11_

-37-

1 to an element of Ci(U) C HB(U) P,00 0 2 A f e HO(U) . q.e.d.

Therefore we can extend Therefore

(p.)((Z p)?f) 1

Now assume

M

is a compact manifold and fix

f e H5 (M,N) . Tf = (h E HS(M,T(N))Iwh = f)

Let projection

7: T(N) -> N . Tf

where

r

is the

is a linear space --- the

linear structure induced by the linear structure on the fibres of

T(N) .

We shall define an inner product on

Tf ,

making it a Hilbert space. Let of

N

[Vj)

and

'j:

of the bundle (Ui,Ki,Pi M,

Ki

be a collection of coordinate neighborhoods T(Vj) -> Vj X I

T(N) .

where

[U i )

n

local trivializations

Pick a finite collection of triples are coordinate neighborhoods of

is a compact subset of

Ui , and

pi E C(M, I)

such that: 1) There exists

Ai a CE(M, I.) such that

2) Support (pi) C-Ui

and

pi

r Ki

pi

=

2

1

3) U tinterior (Ki))= M 4)

For each

i , there exists

J

such that

f(Ui) c v . The existence of the collection

((Ui,Ki,Pi))

follows

from the usual arguments used to construct partitions of

-38-

unity (see 13, p. 27).

We now define an inner

product on on We now define an inner product v2 : Vi Vi X IRn -> _> ~Rn Mn ng: X ~R"

Let Let

second factor. factor.

Let

2 (P 2*j (Piihh

h iS = hij

by projection on the by proSection on the .

h, h'h' E Ty Tf h, h, h' E T,

~ Ui): Ui): Ui -> IR. n ->Vn

Let Let (3 (j

is such that is that

f(Ui f(Ui)>cv C Vj

,n ,~ Ui): h' _> Hs Ui): ,jUi Ui -) [R"

7Tj(Pih' 2= "g~'j(Pih'

hi j h;j

Tf .

Tf

By lemma 37, 317,

hijs h;j hij ~H E HO(U hij' SO (v,,

~R

(hij, h;j)s (hijt hi)s

so

")

is defined. defined. is

Definition 38: 38: Definition Tf X Tf -> Tf X Ty ->

Let Let

.

Z(hij (hij' , hij) hiJ )Ss

(h,h') (hh')i

.

~F~

( ,

It isis clear clear that that It

)

is aa positive positive definite definite is

bilinear map, is an inner product on bilinear map, so so itit is an inner product on Proposition 39: 39: Proposition

product product Proof:

is a Hilbert space with the inner is a Hilbert space with the inner

Tf

Tf

We must show completeness. completeness. Tf Ty

(hn - h-m (hn ij ij i 13'

Therefore, Therefore, since Hs1(U)

E; HS(U) 0 i

-> h: M Ivl -)

HS(U) HS (U) 0

such that such that

T(N) T(N) ,, by by

h(p) = h(P)

. interior (Ki) p p E E interior (Ki)

L

Assume

(hn) ~hn)

is is

.

all i , Then for for all

hiJ

Tf .

Ty

(0)

Cauchy in in Cauchy

h

be be a a map map from from

hn

n ii ij

> 0 hm m )) -_> O

ij iS

as as

n,m -> -> oo o . nm

exists there exists is complete, complete, there is -> h

h n1i ->

hiJi

-'(f(p),

in

H s (U)

Define

in

HS0(U)

Define

hij(p)) (p>>

for for

-39-

We must show that this is a consistant definition; p s interior (Ki) N interior (Ki,)

i.e. that if hij())

(f(p),

f(Ui,) CV,

.

pijl(f(p),

hn

p e Ui n Ui'

.

Hs

an

-> -

Pi: hin

Support (piPi ,

Ui N Ui

2 ,2

=

(p )

Pi'

Pi'hij

Phi'j

(p )

n

in

on

It is clear that

and

C Ui n Ui'

i

( Ui

n U fi

.

)

n Ui ,).

By symmetry Therefore

Ki N Ki'

,

so

h

and since

on

Ui n Ui

well defined.

h

agrees with

p -> ? l(f(p), hij(p))

on the open set interior (Ki) ,

h s HS(MT(N)) .

w2 Pj(Pih) r Ui = hij , so

in

Tf

and

Tf

Proposition 40:

Also

is complete. The topology of

the particular choices of Proof:

hn -> h

q.e.d. Tf

(UiKi, i)

is independent of and

(Vj,ij) .

First we recall that to prove two Hilbert space

norms equivalent, one must only show the identity map is bounded as a map from other norm.

1 1II

Tf

is

S HO(Ui N Ui,)

(f(p)),hi,j,(p))

wh = f

7 2 %jhn

Therefore by lemma 37

.

H(U Hs

in

= 1

)

, if

hn,,(p))

PihJ H s = Pihnij1

(f(p), hij(P)) =p

pi, Pi

-1pi(f(p),

ipi'i' n-n =

Pi'hj

and

, where

hij(p))

To show this we note that for all

function on

(since

(f(p),

=

'

with one norm to

Tf

with the

.

But

Now given any two collections

and

[(Uj 1 , Kilpi,),

refinement

(V ,~f.,))

((Ui,Ki,Pi),

(Vj j))

there is a common

((U i n Ui,, Ki n Ki',

(Vj n Vj,,1

PiP,),

(Pvj V)iV n )

We shall show that the norm of the refinement is equivalent to either of the original ones. (

First we point out that the norm Ui

depends on the chart of so

H(Ui)

is really

Hs((-l(U))

(so:

01 (Ui)

0

is (by the

on

->

Ui )j

a$-lo so: H(O

- 1 (U.1 ) -> Ui

linear isomorphism.

is an is the

(01Ui))-> H(O 1(U It has

so it is a topological

,

HO(Ui)

Hence

M

and

a-lemma 37) a continuous linear map. a 01 o

H(O(Ui

1)

-1

However, given two smooth maps

a continuous inverse,

_

where

:

open set of Euclidean space and chart map.

)

is an open set of

Ui

.

,

is well defined as a

"Hilbertable" topological linear space (a space whose

no

topology can be defined by an inner product). product (( inner product the inner the

,J ),) thaton on

H(U (V,) H sO norm

Therefore

is up to to is defined defined up

equivalence. topological topological equivalence. any that for remark that we remark Next we for any Next constant is aa constant is This This

is is

e HSO HO(Ui, (Ui' j ~) r,fi E

there ,, there

M (hh)of openC (fih, S fi (hh)s (r ihJ rih)h)s

that chart, fi such that

the where the 30 where corollary 30 of corollary consequence or direct consequence aa direct

multiplication. is scalar scalar multiplication. map is bilinear map bilinear (UiKipi) , we replaced (UjrKjrPI) J We replaced (Ui'Ki'Pi) ' we replaced

L

)

pi Pi Pi

by by by

if in in Hence if Hence

p1 2 , the identiy map map pi2 , the identiy pi 2 , the identiy map

_ ~L-~_.-~

I

1

-41-

of

Tf

with the new norm

into

Tf

with the old norm

would be continuous, and therefore an isomorphism. Now show that that the the topology topology of does depend Now we we show of Tf Ty does not not depend on the local trivializations trivializations *, on T(N) T(N) . Let *j and

II 111 1I I

$j be two such trivializations. trivializations.

Let

on

(Vj~*j)J

Tf

induced by: induced

be induced induced by by be

1(Ui.,KiPi),

((Ui,Kip2), C(UIJKI~P 2i

(VAj)}

equivalent to to that that induced induced by by equivalent II II II, 12

shall shall show that that

be the norm

II

1[11 112 112

and let a.nd let

which we know is which is

[(Ui,Kipi), [(UI'KIJPI)'

(Vj,4j)] (Vj,~j)l

.

We We

is is bounded by aa constant times times

~1 II ill·' Note that

72 o

Note that

s2

o

j o Pi 2 h

~j o pi 2h

Ui = 7T~(

t~ Ui

2j aJ

from

V VJ

L(lk nn L(I~

to

dimension of the ffbre the dimension fibre of

R n) ca n>

T(N) T(N)

2o

where

o

a aj

is smooth, smooth, is

and therefore therefore by by lemma lemma 37, 37, and

(

)

and

)2 (( • , )2

giving rise f~iving r~se to to the the norms norms

hk-

where where

nn

is is

(see 13, 13, p. (see p. 35)· 35).

o ph

(pi(jo

2j enoh e a,

oh

= aj of of

(P, (Pi taj (tj

O o a7

h)).(

is is

o~

O Pih)

pih)

H HSs

pi(2j v oo h) h) E HS(Ui,L( In ", Pi taj oo ~I~ E HSO (UI'L( IR

inner products products over be the inner

I1 1111 II1

and and

11 112 112 11

By 30, there By corollary corollary 30, there is is aa constant constant on

defines a

L( tq L( t nJ n,fF~ n) n)

linear transrormatfon linear transformation in in Since

1 Ui . r Ui'

defines a

3

2 0 -1 = ( Pi(aj o a oo h))·(820 2 T2 o ~j o 1CI j o 1C/j O Pi h n: T(N) 7: T(N) -> -> N and srgniries "." ~he action of the signifies the

Also Also

Let

Qj

0 P Pi 2 h o 2h

1) o

--1

j o ?-1 -1

By the the definition definition of of aa vector vector bundle, bundle, By smooth map

o0

o hh f`Ui), r Ui),' Pi(lj i(Aj ao

U Uii

respectively. respectively. C , depending Ci depending i

nv oo hGUi))s hfUi)) s

such that that

n,fn)). IR ")) .

-42-

(hh) 2

C2(h,h) 11 112

From this it is clear that

z Ci

11111

II 1il and

and that

,

is bounded by

11 112

Now we know that the norm (call it

are equivalent.

II 113)

which

corresponds to the set ((U

i

Ki n Ki

0 Ui , ,

,'

Pi Pi'

,

ý Vj n Vj,)]

which we define by the .

replacing

*j,

j

V n v,

(V

II II4

is equivalent to the norm same set with

)

To prove the proposition we need only show that

II 113

is equivalent to

,for

equivalent to

II

II4

v2*jPipi'2 and Also

is

and that therefore all are equivalent.

To show this we replace

form a new

it then follows by symmetry

(((Ui,,Ki,,pi,), (V ,,ij,)

that the norm induced by

)1 II1

pi Pi'

by

pi

2

p2,

to

II I13 2h

r Ui n ui, = (Pi pi'2) (

wv21ji 2 h pi

2

Ui n Ui

E

Hs

( Ui

o *jPih)[ Ui n Ui'

n Ui , ) C HO( U i)

H0(Ui n Ui,) Ci H (Ui)

7 2 o * o pih ý Ui e Hs ( U i

2

and

)

Therefore, since the above inclusions are continuous 2 2 o Pi' Pi h on Ui by corollary 30, the s-norm of v 2 o is less than some constant times the product of the of

Pi Pi'2 ii

and

v2?Pih 2P.

on

Uia .

s-norms

Therefore, there is a

- -

---I

-43-

new constant new constant

C C

(depending (depending on on

pi Pi') pi22 Pi

such such

II

that that

II

II,

isbounded bounded by by cC IIII II,11 . The The proposition proposition follows. follows,

is

Remarks= Remarks:

1) 1)

Note that that in in the construction construction of

not used used the the fact fact that that not

T(N) T(N)

space

space

T(N) T(N)

N N .

Hence, we Hence, if if we

E get a Hilbert E ,, we we get a Hilbert

with with any any vector vector bundle bundle

Tf C HS(M,E) . Ty c KS(ME)

2) 2)

If we consider consider the the special special case case If we

f Identity map, then r = Identity

N N

functions. functions.

= M M and and

Tf T f = ~h (h E HS(M,E) E HS (ME)lrh wh = Identity] Identity)

Tff , ' then, then, is is the the space space of of sections sections of of

Hs HS

we have

is tangent bundle, is the the tangent bundle, but but

only that that it is only it is aa vector vector bundle bundle over over replace replace

Tr Tf

E E ,, which which are are

Palais also also defines defines aa Hilbert Hilbert space space or of such such Palais

functions which which he he calls calls functions

HS(E) HS(E)

(see 17, 17, p. p. 149). 149). (see

We We

shall show than is his. shall show later later than our our definition definition is equivalent equivalent to to his, Derinitlon

41:

Definition 41: CkT

CkT,

k Assume

rf

is is

= (h s Ck(M,T(N))lwh = f) (h e Ck~MT(N))lsh

Ck

.

.

CkTf

Let

CkT,

is a linear space is a linear

with operations defined the same same way way as as for for with the the operations defined the

II IIj be be a a norm norm on

CkTf

defined by by

Ih Ilhll

space

Tf TT .

Let Let

= Z //max CIDah,,iq (Iah Imax D h )j C

i alPEK la (k i

lalk where

Ki Krt

and

hij hij

Hilbert Hilbert structure structure in in

Proposition 42:

T Tf

CkTf

CkTf

p

are defined a.s as in in the definition definition or of the Tr Tf

.

is a Banach space whose is a Banach space whose independent of the choices for for K i and hij independent h 13 ,' Proposition 42:

i

topology is topology is (as in in the las

situation). Situation). jf

`1

r"

-44-

Proof: Tf

One can use essentially the same arguments as for

(see 1, pp. 14-17). C1Tf D Tf

Proposition 4~: Proof'

and the inclusion is continuous.

The inclusion is obvious, since we have picked

large enough so that all

Hs

by the Sobolev lemma for each

on

HO(Ui)

and

I hijll

from the definition of

II

hij

I

> max (IDphij

where

= max (ID hij hij

.

Also,

i , there is a constant

Ilhij l I < Ci II hij IIs

such that

C

functions are

s

II IIs

Ci

is the norm

+ Ihij(P)I)

But

it is clear that

+ Ihij(P)l

.

Therefore

p EK.i 2 1

lal_ N X N

Let

fl(f

is smooth on

has been chosen so that 8 .

in a neighborhood of 9

derivatives on

E

and

Hence

E(Q)

E -1

and E

n

f,(9f,)).

E(X) = (7(x), e(x)).

by

o Pf(h))(p) = E-l (f'(p), eh(p))

(fl

f(@f)

We note that

are smooth maps

and

E- 1

have bounded

respectively, and

e

also

9 .

has bounded derivatives on

The remainder of the proof falls in the context of a general procedure for determining smoothness of maps on Tf .

spaces such as

Therefore we now discuss this pro-

cedure, which is essentially a globalization of the c-lemma. Lemma 46: sets in

M

(Vjj )

on

N

II

product where

< >s

and functions

(Vj II1,

( , h

be a collection of

such that there is a collection of pairs

((Ui,Ki pi), norms,

(Li c Wi c Ki cU

Let

))

on

= pi

Tf h

(h,h') hij

)

is the inner product on

Then the norm

equivalent to

II

Tf .

on

by Ui

,

((Wi,Liai),

II 112

and

)3

and

(pi)

=

3

(1i)

on

M

(Vj.j)}

define

Define a new inner =

i(



we know that there is a constant

The lemma follows if we let

Procedure 47:

(Wi)

Let

h' s Tf' f'

Q

Hs (Wi') $

CO

Qi

if

in

do the following: ,

vI

***

of the norm on exists.

.

v)

respectively.

Wi

by

hi ,and

o h'

by

h'

Tf

in

Qi

For

for

i

and let =

Qi

from

: Q -> Tf' .

(hilh E Q Hs

into (W i )

implies

4

To check that

exists for all

Tf, , we know that

uous linear map from

hh-

Tf'

Make sure that for

Check that

corollary 15,

t

be sets of the above

HS(Wi')

4 Ck-1

Assume

W i>

has the subset topology of

(hn)i -> hi

$(hn)i -> $(h)i

dk$(hi

j

hi -> $(h)i

, where

is

o h

be an open set in

Consider thelmap



Ci

(Wi' and

v2 o

we denote

Let

W, •j hw

i

Tf

v 2 o0j

we denote

such that

C = min [1

and

form defining norms for h e Tf

Ci

Wi>s < Ci goh

by

Proof:

.

To show

existence of c-lemma.

d cg(h,v)

we use steps one and two of the

j,4 o g o

= D(phi(p))

-l1(p, hi(P))

o

2

d co (h,v) = Dh g(v) ,

c

Assuming

,

k-1 Ck-1

we note, using the

v

continuous and to get the

The same estimates work because for any

d (g(h,v)(p) Hence

...

and

and equals

is

*..

to

Hs (W.)

1

v)

vk- (h,vk) which exists

... v- ) .

Therefore, by corollary 15,

any element of v o

j o v

Vi

be bounded in the

~

from

Hs (Wi) , but it is true for Wi ,

CO

sense

d cw(hi ,v)) . This is not true for

v

vi

if

Tf .

Now it is easy to show the smoothness of h

wg

is smooth, because the estimates of

(to get existence of

If

k-1 1 ) = Dh g(v ...

c-lemma (step 6), that

c-lemma require that

=

k-1

We cannot say that the map induced by

H s (Wi)

vi

v

C

Remark:

the

vi(P))

Dh gg(v) = Dh(g)(v)

k-11 1 D~ g(v 1** h

vk) = d (ck-1 Dk (v1

p E Wi

Therefore

o01)(o, g o

so

D -g(v Wg

smooth.

hi -> Wg(h)i

Wmc(h)i(P) = 72 o

dk

.g is

Then

cog: Tf -> Tf,

is in the domain, h i ->

(f1

o *f)(h)i

-l f o is the

.

k-1 vk-)

,

-50-

hi

composition -

and

= w 2 *j

(f',eh) -> w2 *

j 'E-

(f

',

2 *jh)

7 2 %jh) -> e *Jl (f,

(f,

h ->

eh) .

4j ,

T72'

The maps

= eh

E- 1

e,

are smooth, so composing with them is a smooth

('

Also, the maps

operation.

eh -> (f',eh)

and

hi -> (f,hi)

Since the composition of Gateaux

are clearly smooth.

derivatives is the derivative of the composed maps (see H s (Wi')

lemma 14) all Gateaux derivatives exist in can let

Wi

= Wi

'

I

d(*

o

here).

(we

Also

f)(h,v) = D(f',eh)(0, Dhe(v))

and

the corresponding formula holds for higher derivatives. -1

is smooth.

o f

*l

Hence, the procedure 47 applies, so This proves proposition 45. Now we know that that for any

is a smooth manifold such

HS(M,N)

f

f E HS(M,N), a small neighborhood of

looks like (i.e., is homeomorphic to) a neighborhood of

0

in

Tf .

We restrict to the case

define .5 s = (f . H s

(M , M )

f

M= N

bijective and

and

f- 1 6 H s (M,M)L.

We shall show that this set is a group under the operation of composition of mappings, and that it is an open subset of

HS(M,M) , and therefore a manifold.

Lemma 49: p EM ,

___A

Ln

If Tpf

f s

V

s , then

f

is

C1

is a linear isomorphism from

and for all Tp(M)

to

Tf(p)(M)

-51-

Proof:

f

and

f-1

Sobolev lemma.

are

Therefore

1 : T(M) -> T(M) Tf-l

by the remark following the

Tf: T(M) -> T(M)

p e M ,

Hence, for any

-1

T f

are linear maps which are inverse to each

Tf(p)f-1

other, so they are both isomorphisms.

q.e.d.

This tells us that all elements of

Remark:

and

are well defined functions and are

inverse to each other. and

C

.

s

C1

are

homeomorphisms with

C1

Corollary 50:

is closed under composition of mappings.

Proof:

s

1

inverse.

f,g e . s

We shall show

implies

V

f o gs

s

by passing to the local situation and using the composition lemma.

Fix

p s M.

Pick coordinate neighborhoods

of

p, g(p)

and

U, V, W

V ,

of

such that

p

neighborhood of

of

f(p)

and

pf - HO (V1 ) since

U0 C U f(p)

g E HS(UO) .

and

p rV

and

g- 1 (V1 )

g(UO )

so there are neighborhoods

p E Co(V1, ~)

g-1(V 1 ) c U0 .

(pf) o g

Then there is a neighborhood

f(V) C W .

g(U)

f o g(p) , respectively, so that

0

such that .

V0O

_

U0

is a V0, VI

V1 c g(U0)

s -1 (V )) g rg-1 (V 1 ) E H (g 1

Therefore, by the composition lemma Hs(g-1(V 1 )) , so

(pf) o gt g- 1 (v O)

Hs (g-l1 ( 0)) but

__M

pf = f

on

V0

,

so

f o g •g-

(V0 ) E Hs (g - l ( VO))

,

-52-

Hence f og is

HS

g-

HS, s

off-

is

at

p , so

f og

(f og)-l

is

HS

Similarly·

.

HS , and

is

f og E ~5 q.e.d.

Remark:

This tells us that

S

.~

is a group under the opera-

tion of composition of mappings. Lemma 51:

If

C1

is a

f

is a bijective

map, then

Proof:_ We assume on and

s

1. areaC

N

f

induction.

f

Therefore,

f 1l

Therfore

H5 (M,Iv)

HS

C[s/2]±l

is also

ol

map with

C[S/'2]+1

[S/2]±l

and

Now use

.

C1inverse.

(see 13, p. 13). [s/2]+l

is large enough

so that the statements of the composition, alemmas hold. Consider

f

We know that

D(f l) 1

is the map

the composition of i

is

H

o

and

s2±

-

)=Df)

f((f q

->

f1

f 1 (q) -> Df

is smooth and

fore ,

.There

(D ~1 f) f (q)

->

.

non-singul ar .

i

Hk

(D -1 r f (q)

and matrix inverse (call it Df

is

S-

Therefore,

by remark 3 following the composition lemma 28, is-

w-

locally with range and domain in a lEuclidean

space.

i).But

f1

Assume

functions

CS~+

functions will be

functions will be

f-1is

and

HIS7 4]

large enough so that

is a

,

sH(NI)sof

Then

H[([s/2]±l)/2]

map in

and

D(f l)

i o Df

is everywhere

Hence, by the composition lemma,

i o Df o

-----------------------------~

-53-

Hk

is

.

Therefore

D(f-l)

Hk , so

is

f-1

Hk + l

is

The lemma follows. Proposition 52: Proof :



is an open subset of

HS(M,M)

& s

By the above lemma, we can consider

set of bijective f

s

s

E

Hs(M,M)

and

Hs

[fn)

maps with C Hs (M,M)

inverses.

such that

. Then for sufficiently large

hn E Tf , such that hn -> 0

C1

in

Tf .

@f(hn) = fn .

as the Pick f

fn ->

in

n , we can find fn ->

Since

Therefore, by proposition 43,

f

, h n ->

0

C1 Tf

in

Lo ok at

f,

and

h

locally:

SLA

fn = ehn ,

Dfn = Dhn e * Dhn

so

with bounded derivative (on

(Dh n

c-lemma, fore, fn (p

fn ( p )

uniform in

in the and

CI

e

9 C T(M))

converges implies

fn -> f

) ->

.

sense:

Dp f n ->

Dfn

is a .

converges.

C1 b

--- da

map with s

There-

That is, for all

p eM ,

D pf , and this convergence is

p

map with

it, in the

function

So, as in the

It is well known (see 16, p. 21) that if C1

C

Cl C1

is open in

is a

inverse, so is any map sufficiently near sense.

C1

f

Hence for large

inverse, so H s (M ,M) .

fn E

q.e.d.

n , .

fn

is a

Therefore,

· · ·:

-7

ppop-

Remark:

9 s , inherited

We now have a manifold structure on

from the structure on

HS(M,M) .

.V 8 is a smooth manifold and a group; the question arises whether the group operation is smooth.

This we shall

not prove --- it seems that it is false (7) ---but we will obtain some lesser results in this direction. Proposition 53:

f -> f o C . Then

to be the map Proof:

f

Fix

ard charts of

se

and let

Z s at

f

and

0

in

open neighborhoods of -1 4f o R

Then

ft

o *f tRr

The map

Tf

to

ft Tf

and

(!-

and

Uft

are

respectively.

h -> e o h -> e o h o

is the map

We must show

Io

R 0o

AC: h -> h o C is a linear map of

a-lemma.

(Ui )

Pick coordinate neighborhoods 1

(Ui)]

pi, ai,

covering

M,

are also coordinate neighborhoods,

and such that there are sets

functions

TfC

For this we shall use the

continuous.

Uf

where

be stand-

Therefore, we need only show that it is

TfC .

such that

is smooth.

(pf, Uf) (?f ,Uq )

= h o 1 .

-> E-1(f o C, e o h o r)

to be smooth.

R

s

s R :C ->D

and define

C esZ

Let

and pairs

L i c Wi c Ki

Ui

and

(Vj,*j) , which satisfy the

properties of lemma 46, and also the following property: For each

i'

s CO(-l

i , there are functions

(W))

so that

pi' s CO(-1(U))

((•-l(Ui),

1- 1 (Ki),PI)

and

(Vj ,j))

f

-55-

(((-lwi), -1 (Li), ca),

and Tf

.

(For the construction of

11 I1',

11 11 2'

II 113'

and

We shall show that

A : T

II

on

II1

11113'

and

on

11 113

TfC

on Tf

as in lemma 46.

is bounded in the norms

TfC .

and

see 10, p. 3.)

and

-> Rfoc

Tf

define norms on ai

pi,

II i11, 11 112

Now we get norms and

(Vj, j))

By the definition

of these norms, it is enough to show that for each

map

defined by

Ai

in the norms of

h

h

Hs(U1)

o C

1 C-

(w)

HS(-1 (Wi)) .

and

i, the

is bounded But this map

is the composition A -i Hs(Ui) -- > H(C(-1(U))

map

a(,

-1(ui))

(B(f) = f

r

and .

-1(Wi))

a-lemma, and

B

B Hs(-i B-> H -l(wi))

B A

where

A

is the

is the restriction map.

is continuous linear by the

is because it is norm decreasing.

The

proposition follows: Proposition 54:

Let

C edD

.

s

L (f)

(i.e.

C E ) s and

C

is

a smooth map). ->

LC: :

Define

V

by

= C of

.

Then

LC

is

a smooth map. Proof: Z

=

Fix -1

of

origin of

o L Tf

f s

MMIMMý

We look at the map

o 'f which we claim is defined near the and into

GCf , we know that

II

s .

Tf .

h E Cf

From the definition of

if and only if there is an

s > 0

_

_··_

·-

·

-56-

1 l

9,,

Since

was defined using normal coordinates, we know that for

h , the distance from

any such

the length of Since a

< s .

p s M,

such that for all

p to

eh(p)

h(p)

M

r

is compact and

is continuous, there is

such that if the distance from

6

is less than

9f

length less than

6 . We claim

Z

@f .

any

M , the distance between

in

q

in

C(p)

M

and

is

C(q)

so that it contains only vectors of

restricted

is less than

*-l oLL

to

E .

Restrict

p

p

6 , then the distance between

less than

is equal to

a .

o If(h)

h E @f .

Take

But

L

is defined on the

Then by the above, for

C(p)

and

o kf(h) = Ceh .

Ceh(p)

Therefore,

E-1 (f, Ceh)

is defined and it equals

To conclude the proof, we need only show that Zf: 9f -> T f is smooth.

This of course, follows from the -1 general procedure as in the case of *f1 o df , because we know that

E-1 , C

and

e

are smooth maps with bounded

derivatives. Corollary 55:

For any

E•s

,I

L :S

_

s

is

continuous. Proof:

Z :

f -> T f

as above.

for checking continuity.

M

We recall the procedure

hi -> ZC(L)i

is the composition

MOP"-57-

1

3

2

hi ->

(f,h

i

) ->

ej1(hi)

= eh --

> (Cf,Ceh)

4

-,E-1 7T*JIE ((f,leh).

-- >

Here all steps but Tp

3

are clearly continuous.

is everywhere non-singular and H

means

hn

> hij

f [s/2]+l(w)

.

in

H(Ui)

so

But

Hs (W)

hni- >-> hi hi inin hn -> h in

Therefore, by the weak

continuous, so the sequence is also.

e-lemma

3

is

This proves the cor-

ollary. i: 4s

Proposition 56:

V s by

_>

1 ->

Proof:

8

Since

show that

i

S' I = * the map

and

L

continuous at -1

o i o a*

are continuous, we need only Id .

But in the chart

has the following form:

p -> E-l(eh(p),p) .

usual reasoning.

.

I(h)

This map is smooth by the

The proposition follows.

This concludes our discussion of We shall now look at

Riemannian metrics on

001.

is continuous,

is a topological group.

so • s

of ~

-1

HS

s ,the

functions and space of

HS

M .

I

M

-----------M

-58-

IV.

The Space of Let

StT*

t-tensors over

Hs

Metrics.

be the bundle of symmetric co-variant M (the symmetrization of the cotangent

bundle tensored with itself Hs

the Hilbert space of Tf

where

StT*). of

Ck(StT *

StT*

)

(defined as

Definition 57:

The

times).

sections of

f = Id: M -> M

Let

t

and

T(N)

form on

y(p)

HS, respectively

Ck

is replaced by

Ck

sections

Ck,

respectively

Hs , respectively

y of

S2 T*

Ck

such that for

Tp(M)

metrics

Lemma 58: Proof:

(defined as

defines a positive definite bilinear

We shall call the set of of

StT*

be

CkTf).

respectively smooth, sections p E M ,

Hs(StT*)

be the Banach space of

smooth Riemannian metrics are the

all

Let

Hs

metrics

s , the set

Ck9?" and the set of smooth metrics

C0 9~7

is an open subset of

9

.

CO(S2T*)

We use the fact that the set of positive definite

symmetric matrices is open in the set of symmetric matrices. Let of

M ,

Ki C Ui

[Ui] (Ki) and

be a finite set of coordinate neighborhoods a collection of compact sets such that U Ki = M .

If

?i:

S2 T*(Ui) -> Ui

k m m

is the natural trivialization induced by a coordinate system

---

I

-~1

-59-

xi

xn ; that is,

***

I((kgjk dxJdxk)p)

(Pglgg129' '',gnn) )

jk

then

(gjk)p

is a positive definite matrix for all

is a continuous function on

gjk

Ui

is compact.

Ki

and

p e U

Therefore, because of the fact that the set of positive definite matrices is open, we can find an for any section

[

max

h

of

si , such that

S2 T* if

2 o *i(h) - 72 o * (g)


-s

II b I= max

114xl=1111

( b(x,y) ).)

=1

5 )) s X HS(S 2 T* ) , B(T(?S

=

B(Hs(S 2 T*))

such that for all

which induces the topology on struct a metric on %P7 s

1

on M

=6Sx B(Hs(S 2 T*))

is a smooth map

is a positive definite bilinear form on

action of O

Tp(X) .

(It is a Banach space

Hence a Riemannian metric on t s P:

is positive

be the set of symmetric bounded

under the operator norm Since

Ck(s2T*)

is naturally identified with

; that is, T( IS)

Definition 61: space.

T (Vns)

,fl= c0oh7n

HS(s2 T*) , it has a

is an open set in

,

Ck

and

The proposition follows.

natural manifold structure based on each

CO(S2T*)

are continuous.

0n HS(S 2 T *

= CO

and Ws

C

HS(S 2 T*) .

ye

s

Hs(S2T*)(C! T7T(q

s))

We shall con-

which is invariant under the

s . To define this metric we use the

I

-d

·

·:·

-61-

HS(S 2T*)

definition of

given in 17, p. 149, so we must

first discuss that definition and show that it coincides with ours. HS (S2 T* ) in

We briefly review the construction of (17). Let

E

E .

the smooth sections of

Let

Ce(M,Pj

Ip be the ideal of functions in p .

vanish at

Zk(E) = Ik+l

*

Ce(E) C(E)

is Let

U

sections of

E

k-jets of

E

1

over

Z pi(x1

polynomials in

fi

) .

Let

E

p .

at

We

Jk(E) , whose fibre at each

is trivial over

(xl

M

U , and let sl ... sm

U . Let

containing x1 ... xn be smooth

U which are a basis for the fibre Consider the set of sections of

at each point.

of the form

where

Ce(M, R

be a coordinate neighborhood of

be coordinate functions on

U

which

J (E)

p , such that

of

Ce(M,IR)

and

k (E) = Ce(E)/Zk(E)

and let

shall define a vector bundle p eM

be

p EM

Fix

is a module over

k (E) is called the set of

Ce(E)

be the ring of

M .

the smooth real-valued functions on let

M , and

be a vector bundle over

*** ***

xn)si

xn) .

where

pi ( x l

..

xn )

It is a linear space.

are smooth functions on

U .

E are

We

i By Taylor's theorem

-62-

we know that at fixed

p

in

for each

fi

fi

p

such that

pi

exists some polynomial

U

k+

pi

Also, if a polynomial

is in

i

there

,

k+l

sE

, its first

Ik

derivatives are zero, so it must be zero.

.

i

Hence

k+l J (E)

is isomorphic to the vector space whose elements have form Z pi

si

To define the vector bundle

k (E)

we need only

specify sufficiently many smooth local sections; i.e. enough smooth local sections to span each fibre 1,x

The elementary functions

1

J (E)

J )2,xlx2,...(xIr ,x2,...,xnfx

span the set of polynomials of degree less than or equal to

k .

Hence the elementary functions times

span

Jk(E) .

over

U .

Let these be smooth sections of

1 2 s ,s ,...

s

n

Jk(E)

It is easy to see that these functions trans-

form smoothly on overlapping coordinate systems, so they define a smooth bundle

Jk(E) .

There is a natural linear injection

If

s E Ce(E) , ik(s)p

represented by definition of

s

in

Jk(E)

Palais defines

ik: Ce(E) -> CO(Jk(E))

is defined to be the element Ce(E)/Z (E) that

HS(E)

ik(s)

It follows from the is smooth whenever

as follows:

Let

s have

Jk(E)

a smooth Riemannian structure --- a positive definite

inner product

< , >

on each fibre such that the product

of any two smooth sections is a smooth function on

M .

is.

-63-

Define

H (Jk(E))

to be the completion of

CO (k(E))

with respect to the norm induced from the inner product = Ips ,sp dp(p)

(s,s') on

M .

p is a smooth measure

where

Palais shows that

HO (Jk(E))

vector space does not depend on the for

Jk(E)

or on the measure

Jk: Cm(E) ->

injection.

as a topological Riemannian structure

P.

C (Jk(E)) C HO(k(E))

is a linear

Therefore, an inner product on CO(E) .

an inner product on completion of

CO(E)

Define

HO(k(E))

HS(E)

is

to be the

with respect to this inner product.

Now our proof of the equivalence of the two definitions will be done with two lemmas. A-Lemma 62:

Ce(E) Hs

the set of B-Lemma 63:

as a map from topology) to

Proof of a norm

is dense in

sections of

E

HS(E)

A-Lemma: II Ij for

COe(E)

as a subset of

Ce(E)

and let

partition of unity subordinate to i , there is a smooth function

Ai

(Palais

OH(E)

((Ui,Ki' Pi),

Hs(E)

(Vj,

j)]

to define

(cai

be a smooth

[Ui)

such that for each

such that

1

ai = 1

(This can be done, for given any partition of unity let

ai

=

(a,)21 Z(a)

2

.

Then

is

is continuous

HS(E)

as a subset of

Pick

HS(E)

with our topology.

The identity map on Ce(E)

where

Ai =

11

i/ 4z(a)2

2 1

a' which

1

i~J(i)2

is clearly smooth since Now pick

h s HS(E) . By lemma 37,

=i(Cr2 o jo h Ui) E HO(Ui) that II hi - ai( the norm on

is nowhere zero.)

2

o

h :M ->E by

Pick hi •0 C0(Ui)

.

Co

is in

i

h0 (p)=

Ce(E) .

is close to

Let

h

p jU .

hO = Zh HS(E) .

in

where Let

ih

hi -

We shall show that

We know

l=

and

h = Zaihi , so

i , hi

II Pk'2 jh 0 - Pk1i

2

is close

jh I % k

IIIk is the norm on HO(Uk) Xik = Pk

2

jh-

Pkaijh .

2

jhi - aijh I"k = Ci E

Pkai±jh

such that

II h i

- aih

Ci , such that For any

.

k,

I Xik lk pkv2 jh0

have their support included in: Uk n Ui '

so they are elements of

Cik

k = i , corollary 30

If

tells us that there is a constant < Ci

1

'1 hp)) p Ui

it is sufficient to show that for each to aih .

Hs(Ui)

II Xik Ik

IZ= 2 II Xik II k

Therefore,

IIh

Ui)

such

Define

S(P.90) hi

oj(aih

-o 0

o h Ui) II < e where II 1i is

HO(U i ) .

0

2

Cik k

and there is a constant

Xik iii .

Hence

Ci Cik

- h I < k C. Cik

, and since

E

i,k

can be made arbitrarily small,

CO(E)

is dense in

H (E) q.e.d.

-65r

Proof of

B-Lemma:

inner product on

A norm on

JS(E) .

HS(E)

is defined via an

We shall construct several

inner products which are "locally flat" on various neighborhoods of

M . Each will give a norm on

of course, all the norms are equivalent. a set

HS(E)

and,

Therefore, if

P is bounded in one such norm it is bounded in all

of them.

We shall show that this implies that

bounded in

HS(E)

Let

(Vk.)

E

'

Vik

2) For each

as well.

M

such that:

is a trivial bundle for each

K

call

Vi0k , such that

UVik

= M 3) The sets

Uk

((Uk KkPk), (VJ j)) *j

over

Vio k k Fix

Uk which we shall

Vik

Uk 0 Vik = 0

if

i

10

and

can be used to form a collection

which defines a norm for

HS(E) ,

is the restriction of the trivialization of

k ,

subbordinate to

and let

(ai)

E

be a partition of unity

(Vik) , and let

be a trivialization of basis of

Vik

there exists a neighborhood

contained in a single neighborhood

where

is

be a finite collection of convex coordinate

neighborhoods on 1)

P

E(Vik) .

'Ii:

Let

E(Vik) ->

Vik X Rm

(el,... em) be a

PK m and sj , defined by sj(p) = t~l(p,e ) be

1

-1

'

`

-66-

E(Vik) , and if

at each fibre of

are a basis

are the elementary

(pi)

Vik , of degree less

polynomials in the coordinates of

(pis])

s , the set of sections

than or equal to

JS(E)

inner product of

over

>i


i


k = Zai

product on

i .

orthonormal basis since Uk

i0

Vik

be defined to be

ai = 0

p = Z ai(dx 1 dx 2

Consider the set it defines on

is

is an

i

if

10

and

s IIo) 2

... dxn)ik

Let a measure on *** dxn)i ..

HS(E)

((Uk,Kk,Pk),

(dx

by


=

Therefore

on

.

Bk(S)

'I

p

n

in

fUkla<sa((D

p

Vki0

f))2 dx1

dx n

...

Uk , and

IIS 11k where pk

k IIs 110k

=

max

IaI.s

is the norm of

We see that it is bounded by

The lemma follows.

--.N

.** xn

Uk .

2 2 so (IsIIo) • -k CII SS I k HS(E)

p

are an orthonormal basis on

p = dx1 ... dxn

in

2

for all

taIL~s

II 2k = f M

2

E IS

k

4 Jk 0 k

S

II S Ilk

· ·_·

·

-681

---------so

Proposition 64: Proof:

HS(E) = HS(E)

By the

and by the

A-lemma,

as a topological vector space.

Cw(E)

is dense in each space,

B-lemma, the identity map is continuous for

the two topologies.

Therefore, it extends to a continuous

map (also the identity)

id: Ha(E) -> HS(E)

this map is

injective, so to show isomorphism we need only show it is onto.

Let

Uk, Vik

be as in the

B-lemma, and let

be a partition of unity subordinate to

ak

is the square of a smooth function.

Then ak

r2 o *j o h l Uk

o h) t Uk e HO(Uk)

o

2

[fn )

Hs

is an

Co(Uk)

H(Uk)

Let

such that

p l = P)"

->

II

Ilk

so

k(

by lemma 37.

on

Hs(E) , so

h s HS(E)

.

Uk , so

Let Uk

'

in

by

P E Uk

jh) t Uk ,'in HO(Un

2

he Hs(E)

Let

function on

Then it is clear from the definition of

fn

Uk , such that each

fn -> ak(7 2 %jh)

(h) C C(E)

(ak]

h ->a

ak h E HS(E) .

Therefore,

I1 Ilk

But

in

h = Z akh

id: HS(E) -> HS(E)

and hence it is an isomorphism.

that since

k

is onto

q.e.d.

In order to define and discuss the properties of the Riemannian metric

P: ~Is -> BO(HS(S 2 T*)) , we shall need

to use differential operators.

We now define such operators

-69-

and mention a few of their properties.

Details can be

found in 17, Chapter IV. Definition 65: M .

E

and

F

D: Ce(E) -> Ce(F)

A map

of order

Let

k

s E C(E)

if

D

be vector bundles over is a differential operator

is a linear map such that if

s E Z (E) , then

and

D(s)

= 0

Jet bundles are a "universal space" for differential operators in the following sense: is a

kth

D: jk(E) ->

F

map

Jk: Ce(E) ->

such that if

is the natural injection, then D

D: Ce(E) -> C'(F)

order differential operator, then there is a

linear map

this,

If

D = D * jk .

C (jk(E))

Because of

can be uniquely extended to a continuous linear

D: HS(E) ->

D: Jk(E) -> F

HS-k(F)

D: Ce(E) -> Cm(F)

the map

D(s) = D o jk(s)

s > k .

for all

is a

kth

Also, given -defined by

order differential operator.

We note that a differential operator of order a differential operator of

k+i

k

is also

for any positive integer

i , and also that the composition of two operators of order

k

and

I

is

an operator of order

k+1 .

We give several examples of differential operators which we will have occasion to use: Example 66:

Let

V

affine connection on

----..

: Ce(T(M)) X Ce(E) -> Cm(E) E .

That is

V

be an

is a bilinear map

-1

-70-

such that if

then

, V E Ce(T(M))

f e Ce(M,R)

V

VfVS = f

S E Cc(E)

V (fs) = (Vf)S +

and

S

,

D: C (E) -> C(E X T*(M)) by D(X)(Y) = D

V

f

,

S

X (X C'(E), YE C M ))

is a first order differential operator called the co-

variant derivative (see 17, p. 85). Example 67: Y7

7

Fix a Riemannian metric

C~(S 2 T*)

.

2 T*) a: C'(T(M)) -> Cm(S

Let

7

a(X) = QX(7) , the Lie derivative of

y = gijdxi

If in local coordinates

then

(see 9, p. If

k av

he Xk +6 +

6- k

= (vk ki

@X(Y)

i gr

Then

be defined by

with respect to

dxJ and gkdxx kv) dxi

X = vi

X. i

6D dx j 0 dx

kdx

'

107).

X s ZP(T(M))

Therefore, Example 68:

a

,

v

i6

is a first

= 0

symmetric with respect to

a subbundle of

so

y as in example 67, and let

Fix

and

=

and

X,Y

a(x)

=0

order differential operator.

bundle of homomorphisms of

A s SHp

M .

on

T p(M) ,

T(M) y ;

into

SH

be the

T(M) , which are

that is, if

p EM ,

yp(AX,Y) = 7p(X,AY) .

Hom(T(M),T(M)) = T(M)0Q T*(M).

SH

is

We shall

construct a first order differential operator

P: Ce(SH) -> C(S2T* T(M)) . Whenever we use local coordinates we assume that repeated indices are summed.

1



-71-

We first recall that the exponential map on matrices (call it exp), takes the set of symmetric matrices onto the set of positive definite symmetric matrices and is a diffeomorphism between these two sets (with their usual The exponential map can be applied to

manifold structure).

any linear transformation on finite dimensional vector spaces, so it induces a map

exp: Hom(T(M), T(M)) -> Hom(T(M),T(M)).

exp(SH) C SH , and exp t SH

is a smooth injective map onto

the positive definite elements of

SH , where we say

Let Then

is positive definite.

yp(AX,Y)

X,Y ->

c(SH)

s

' * a

and define

M .

Cm(S2 T* ( T(M))

7

y

li,

because

and

and V a 7 * a

is a

be the Riemannian

respectively,

to be the difference tensor of

; that is, if X,Y E O(T(M)) , -

p(s)(X,Y) = Y

ap = exp(sp)

by

for the definition of 17 ). We define

(see 11, p. 71

and

1

Let

connections associated to

7a

a E CC(SH)

(7 * a)p(X,Y) = yp(apX,Y)

defined by

Riemannian metric on

p(s)

defined

Tp(M)

is positive definite if the inner product on by

A E SHp

Va

VXY X.

and

V

(s)

is symmetric in

are symmetric connections

and

X

(see

p. 64). We now show that

tial operator. Sgi s ~

Y -gg,

P

is a first order linear differen-

To do so we use local coordinates.

ssP

a ~ ai

and

k

ij

and

Let

a 1

ij be

In general x ~ xij means that the tensor field is expressed locally by the functions (xij)

X

-72-

the Riemannian-Christoffel symbols corresponding to and

Va

From

the Levi-Civita formula we know: -k- ( g

rk

ij

+

)

where

for

Srk

y

g rj

Substituting

aa 2 rgrgi aa ,xjx 21

SXi

2

b

is the matrix

we get

ksb a fVa a - 1 gbs

15

gki

x

gkj , (see 11, p. 72).

y * a ~ air gr

where

g -i xj

g

inverse of

(+)

p(s) ~ ar k iJ

Then

respectively.

V

is the matrix inverse of

computation we get:

.

a

rj

By direct

computation we get: a

k

1

k

Pij

ij

ra

Ia

ks bS

i

(x

ýa. r

ýa r grj

+

x

j

-I grj - ai

gri

ax aaiJ Note that

8(exp(s))i

(ap k

if

k ax

(siJ) p = O

1

_

= exp(s)i

k

-

Tx k

Therefore

j

siJ i

and

s

i

6

=0,

ki

x

k)

i) 6x

1 (gks 6 6r ~ -x = 0 .

gkk

i

P

))

6gJ px p

1 0

)

=

-73-

P is first

Hence

jl: Ce(SH) -> OC(J 1 (sH))

where

P(s) = P o jl(s)

order, and and

P: Jl(SH) -> S 2 T*

is some bundle map.

P is

B is linear we have to show that

To show that

linear or that for any

p

M , the map

in

M . Sp is is linear. Fix some p P : Jl(SH) -> S2 T p p p a symmetric linear transformation so it is conjugate to tiJ =

a diagonal map --- a map

.

6

Hence we can

choose a coordinate system so that at i p ,

With this coordinate system, p .

at a

k

1

ij

ij

2

From this, it

+

6 6 t

t

is clear that

P

kg(

cannot say immediately that s

P

s'

and

j

bi

= e-

ri

r9

6

Jp(SH)

r



gij

maps zero into zero

P(?s) = ?ý(s)) .

and is homogeneous (i.e.,

know that given

and

s r

6sr

k

x i grj

I

6i

i

p ,

Hence at

k

= e

ai

6

si=

However we

is linear, for we do not

there is a single coordinate

system for which they are both diagonal.

But we do have

the following:

Lemma 69: and

If

f(O) = 0

a

F

then

map f

f:

IRn _> km is homogeneous

is linear.

-74-

Proof:

Fix

asI~n

m = 1 , or

and assume

f

is real

valued. f(a)

-•(f(?a))

=

By the mean value theorem,

(ha) = ~( (0) + Dt a f(a)) Let

->

Since

D0f

0

we get

D 0 f(?a) = D 0 f(a)

f(a) = f(0) +

is linear, so is m ) . E:ch

f = (fl,f 2 ,..

0 1 , we let

is linear, so

f

is linear. q.e.d.

(+) and the fact that

From the formula smooth it is clear that the above lemma,,

1J

p

"exp"

(SH) is smooth.

is linear so

P

is

Hence by

is a first order

linear differential operator.

p: 9Z-> B(Hs(S2T*)) each

'y6

an inner product on

to show how this is done. bundle

E

JS(S2T ) .

We proceed

First we note that for any

and any positive

of bundles where

is defined by assigning to

0 -> SkT*Q Eik

k , there isan exact sequence k(E)

>

k-l(E

)

->

0

jk

is the natural projection Ce(E)/Z k (E) -> CO(E)/Zk- 1(E) Sik p p induced by the inclusion Zk(E) C Zk-l(E) and - p ik: skT*

Given

e e E

s .em(E)

L-

E ->

Jk(E)

and

such that

is defined as follows:

t E T*(M) , pick sp = e

and

f

C~m(M, I )

dfp = t .

Then

and

-75-

ik(tk

Sp)

is the image of

For the proof that

fkss e C(E)

ik

in

Jk(E)

is well defined (i.e.,

independent of the choice of

f

and

s) and that the

sequence is exact, see 17, p. 58. Proposition 70:

D: Ce(E) -> OC(T*(M)@ E) D: J1 (E) -> T*(M)d

derivative, then

is a splitting of

E -> Jl(E) -> JOE = E

T*(M)4

the exact sequence

E

is a covariant

and

conversely any such splitting defines a covariant derivative. Proof:

See 17, pp. 84-85.

Proposition 71: for bundles

E

If D1

and

are covariant derivatives

D2

F , then

and

D: Cd(E ( F) ->

C (T*(M)

D(e I f)

f + 7 1 2 (e QP D 2 f)

= D1 e e

) E aF)

defined by (where

71 2 : E @ T*(M)@ F -> T*(M) (@E0F by interchanging T*(M) and E factors) is a covariant derivative for Proof: Remark:

D

'

E@ F

See 17, pp. 87-88. If

and

E = F

C (S2E): C (S2E)

D 1 = D 2 , then

-> Co(T*®S2E)

and

D

r

Ce(S

2 E)

is a covariant derivative. Corollary 72:

Given a finite set of bundles

each with covariant derivative derivative

1

D

on

E1 ® Ee2

(Ei1 =1

Di , there is a covariant *. 0 Ek

such that

k

D(el@ e2.''*

@en) = Z

li(el®

e 2 @0 ..

i=1

where

vli

Note:

The above

on

on

(T

S2T

)

D

will be written as

S2 T* .

Ak = Dk o Dk-

t h

factors.

D1(

D

on

8 ek)

D2

... ·

T*

Ak

Let

be the

1

o ...

o DO '

S2T )

Dk

)

order differential

defined by

induces a linear bundle

k

S2 T*.

Ak: Jk(S 2 T*) -> (T*)

kth

Dk

and

we get a covariant derivative (call it

,

operator Ak: Ce(S2T * ) -> C'(T *k

map

i

interchanges the first and

Now given covariant derivatives DO

thDiei.

Let

Sk: (T*)k -> SkT*

be the canonical map which "symmetrized" tensors: Sk(t

to(1)ZI

1

t

is the symmetric group on Let

Proposition 73:

SAk:

the exact sequence

IdS2

k(S2T

)

)

be defined by:

o Ak

-> STkT*@ S2 T*

SkT* O S 2 T* -> Jk(S2 T*) ->

splits k-l (S2 T)

See 17, p. 90.

Corollary 74:

Covariant derivatives on

induce an isomorphism: Jk(S2T.)

k I iT&

i=0

-L-

kI

symbols.

SAk: Jk(s2 T*) -> SkT*@ S2 T* SAIk = (S I(k

Proof:

k

®t(k) where

S 2 T*

T*(M)

and

S2 T*

.

-77-

Proof:

The splitting of the above proposition gives an

isomorphism

Ji(S2T*)

2 T* S2

SIT*"(

Ji-l(S2T*) .

corollary follows from an induction on 7 se¶

i

T(M) ; using it we

is an inner product on

Jk(S 2 T*) .

shall construct an inner product on

)

y defines an isomorphism (also called y: T(M) -> T*(M)

by:

y(V)

The

X ->

is the map

(7(V),X)

By means of this isomorphism it gives an inner product on T*(M) .

by:

Also

(V1

gives an inner product on

y

@V2 @ "'" @Vk,

W1@W2 0

0 Wk) =r

Sk*

and therefore on the subbundle

.

Hence k

inner product

on

S2 T*

SiT*

(T*)k k

and on

Z

i=1 7

1

defined

(ViWi) gives an 2T S3

SiT*

i=0

Therefore, we seek an isomorphism between k

Z

SiT*

S2 T

.

By corollary 74,

Jk(S2T )

and

it is sufficient to

i=0 find covariant derivatives on Given connection

V

V'

derivative on S2 T*

m

v'

on

is defined by

v E CO(T(M)) ,

as well.

T (M)

T(M) , the Riemannian connection.

we have an isomorphism between

get a connection

and

, we have a canonical affine

7y 6 on

S2TT

w T*(M)

T(M)

and

T*(M)

Also

so we

T*(M) . V'm = y(7v(7-1w))

Ce(T*(M)) .

'

where

induces a covariant

and by corollary 72, we get one on

Therefore

7

defines an isomorphism

U,

-78-

E

S2 T* • JS(S 2 T*)

SiT 0

7

dx 2 .'' dxn .

-det(Vij)

y

Therefore

HO(Js(S2 T)) , and this gives

induces an inner product on

HS(S 2 T*) .

an inner product on

))

g: ar -> B(H(S 2T

Let

Definition 75:

.

M --- in local coordi-

also defines a measure on

nates it is

J (S2 T)

and an inner product on

the inner product on

HS(S 2 T*)

equals

p.(y)

by

y in the

induced by

manner described above.

B to construct a ýk

ential operator

C (SH) X C (S2 T*)

operator from

70 E ~

s , and

-h

-

order differential

C'(SkT @ S2T

to

)

be the subbundle of

SH

and let

W

We begin by using the differ-

that the extension is mooth.

Fix

extends to

p

We proceed to show that

Hom(T(M),T(M)) , consisting of homomorphisms which are

SAk: jkJ(S2T

by

0o .

Let

SAk:

induced by

->

Pick

SkT*§ S2 T

as C(SH) .

.

be the homomorphism induced

y0*(exp a)

Then S2 T

k(S2T*) -> SkT* yO*(exp a)

Let

y0 .

symmetric with respect to

Let

.

e

be the homomorphism

SAk,

C (S2T * ) -> Co(SkTD 2T*)

SAk:

be the differential operators defined by these homomorphisms.

Proposition 76: by

Let

Bk: C'(SHI&

Pk(a I& g) = SAk(g) - SAk(g)

S 2T * )

k

->

is a

*) Ca(SkT kth

S2 T * )

order

I

PWO -79-

differential operator. Proof: Case I:

k = 1.

and

T*(M)

Let

7

V

a

k :

We proceed by induction on

induced by the metrics

yo

the connections on

covariant derivatives.

,.

Lemma 77:

Let

Note that

P(a )

Prooof of lemma:

'Du(X,V)

and

T*(M)

T(M) 'D ,

O(T*(M)) SE Ce

.

E Ce(S 2T* I T) Pick

and

Da

y0

and

D(V 0 w) - Da(V

=

so

B(a)(w) e Ce(S2T CCO((T*) 2)

X,V E Ce(T( M )) .

V x(To() ),V)

o(

-•

y .

W)=

0 Pl(a Q V

(V(7 ())

) ) , V)

X) -X(a(V)) + W(V V The lemma follows.

be covariant derivatives on

S2T*

Using corollary 72,

('D -

'Da)(V)@ W +

= B(a)(V)@ W +

Therefore

their

'Dco - 'Daw = B(a)(w) .

Then

= W(P(a)(x,v))

induced by

'Da

x(v,w) = (Vxv ,w)+ (V, V xw)

- 'DaMt(XV)

D

'7 and

We recall that for any Riemannian

= X(M(V)) - W((xv)

Now let

7 = yO*(exp a).

and

and t7 a be the connections on

connection

T(M)

We first analyse the connections on

W) S(sA1

12

Tl2(V

(V@ O

6('D

TDa)(W))

(a)(W))

- SA)(v 0 w)

= P(a)(V)@ w + v12 (V ID(a)(W)) .

---

I

-

~ ~___

-80-

order differential operator it

P is a first

Since

is clear from the above formula that 2 Pi: Ce(SH@ S2 T*) -> C'(T*b S T*))

Case II:

We shall assume the proposition for

Pk+l

prove it for derivatives on SAk

Then

SAk:

" Let

kth

k

k

a

y7

and

70

are the linear maps

S2T

order operators

1

...

k-1

o ...

and

o D1 oD a .

1

) o Ak + S

a

o (Dk+1l k+l k+1

SAk+l - SAk+l

and

Pk

be the covariant

induced by

k(2T*) -> SkT *

Dk Sk o Ak = Sk o Dk a k S o A = S oD oD

a

Di a

and

Di

S2 T*

(T*)i&

corresponding to the

k

is also.

-Dk+

+

k+1

o (Ak

a Dk+l

o

k)

We look at the two terms on the left of the above expression. In the first term,

v1

where and

vli

in case I,

..

rl (vl

r vk

Vk+2)

k+l

'.

v2

(Dk+1 - Da+ 1 ( k+2 Z i=1

**

(a)(v,)

kE C ((T*)k)

Vk+ 1

interchanges the first and a @ vl

) .*

ith

Vk+2 -> (Dk+l

**D

Vk+2

Vk+l

C(S

factors.

2 T*)

As

ak+2 - Dk+ 1

is a first order operator from

C

shl.

(SH

& (T*)k

S2T*) -> Ce((T*)k

+- l

S2T*) , and it is

Vk+ 2 )

-81-

Also

Ak

Hence

S+o

of

-

a

o A

Second term:

k+:

(k+l) S2 T*))

kk ka) is a k t Hence Ak Ce(SH® S T*). -

Hom(Jk(SHOD S2T*),

which gives an element of o

so it is C (Sk+lT*(

to

k S k o (A 2

By assumtion,

induces an element of

Sk+

S2 T*)

Ce(SH

order linear operator on

Also

is defined by an element

Hom(J 1 (SH)@ Jk(S 2 T*),Sk+lT*O S2 T*)

order operator from

S2 T*).

Hom(Jk(S 2 T*),(T*)kg S2 T * )

gives an element of (Dk+

(T*)k 0 S2 T*),(T*)k+1l

Hom(J1(SH 0

induced by an element of

(T*)k

h

k

a

S2T*))

Hom(Jk+l(SH & S2 T*), J1((T*)kM S2 T*)). S2T * )

Jl((T*)k(

->

Sk+1T*

S2 T*

is a linear map. S

Hence element

o

k+1

Pk+l

a

k+1

of

so it is a

(k+l)

C (Sk+1T*

S2T)

o (Ak _

k)

a

is induced by an

S2 T*),

Hom(Jk+1(SH (

Sk+ 1 T X* S2 T*) Cd(SH@ S2 T*)

order operator from ,

which is

(a @ q) -> Sk+ 1 o Dk

o (Ak Hence each term of of order

k+l

Since

Ps

.

Pk+l

to

is

k)(q)

a differential operator

This concludes the proof. is

an

sth

order differential operator,

Ps: Hs(SH (0 S2T*) -> HO(SsT* 0 S 2 T*))

is a continuous linear

map. For any

k < s ,

k: Ce(SH

S 2T * ) -> Cc(SkT*I

is a differential operator of order

1

s .

S 2 T*)

Therefore, the

0

-82-

s

same is true of 1r: Cm(SHOS 2 T*) -> CO( where F

s • Bk . 1=0

defined to be

is

5: HS(SH® S2 T1)

-> HO( Z

S2 T*)

Z SkT*

k=0 Hence

S2 T * )

SkT*I

k=0 We note that the map

9

t:

f-> B(Hs(S2T*))

depends

on three things: 1)

The isomorphism

IkT

Js(S2T*)C -t)

T

k=0 2)

The inner product on

3)

The measure on

S2 T*

Z Sk T*' k=0

M .

Therefore, to prove that

L is smooth we look at each of

these separately. 1) E:

By fixing

YO 0 E

-s Hs(SH)

*

Since

exp

,

defined by

we got a correspondence 7 = 70* a ->

log a .

is a diffeomorphism from symmetric to

positive definite symmetric matrices, example 48 tells us Therefore

that this correspondence is a diffeomorphism. to show

s

is smooth we can work either on

p

or on

Hs (SH) The standard map

q: SH X S2T * -> SH

S2T * S

is bilinear.

Hence by lemma 29, there is an induced map q: Hs (SH) X H s (S T * ) -> Hs (S H and bilinear.

1

(

S2 T

o q: Hs(SH) X HS(S

) which is continuous 2 T*)

0 -> HO(

kT*@ S2T

k=0

is therefore also continuous and bilinear.

-----

I

-83-

P1 : HS(SH) -> L(Ha(S 2 T*), HO( z SkT k=O

Let

be defined by

1!(a)(g)

o q(a,q) , then

="

a continuous linear map.

S2T)) is

'1

clearly

Also from the construction of '5

we know the following: Proposition 78: h0,h: C (S2T*)

Let ->

a e C6(SH),

Cc(

7 = y 0 *(exp a)

SkT*® S 2 T )

Z

, and let

be the linear maps

k=O yo

induced by

g

and

C(S2T ) ,

Proof:

y

respectively.

1(a)(g)

Then for all

= ho(g) - h(g) .

Simply follow the steps of the construction of %I s -> L(Hs(S 2 T ),

Definition.79:

6 .

SIT*@ S2T ))

HO( Z

k=0 (y)

by

tl

pI

are. 2)

(E-

= ho -

(

) )

p1

smooth since

E-1

and

by the above proposition.

pl(Y)(g) = h(g)

Also

is

We know that since every

y 6

it defines a continuous inner product on

s

is continuous,

T*(M)

and hence

S

on

S2 T* .

Z SkT* k=0

tnner products on

Let

Z

SkT*

PR

be the space of continuous

S2 T*

and let

P2: 9n s -> PR

k=O P2 (7)

be defined by: Z SkT* a k=O

S2 T

is the inner product on

induced by

y .

Note that

PR

is

an open

convex cone in the linear space of continuous bilinear maps

on Z skT

S2 T

k=0 Proposition 80:

P2 :

~

s -> PR

is smooth.

mod

-84-

We follow several easy steps:

Proof:

i:

( 9s

T*(M)

on

s

defined by

y .

where

is the matrix inverse of

gij

HS(S2 T)).

is an open convex cone in

s _>

V

Riemannian structures

Hs

be the space of

Let ?Zs

a)

by

i(y)

is the inner product on

gij .

T*(M) dx j

i(gij dx

In local coordinates

Hs b)

Hs

7

when

is and that the map

be defined by: J2

E.

j2 (g)(V

Let W, V'

0

s (T*)

c)

The map

-_>

d)

s(T* @ ® T*)

W') = g(V,V')g(W,W')

qk:

s((T*)k)

? s(T*k) ->

restricting an inner product on the subbundle s S2 ^ k k: Hs(S2 (T*k

- >

A: Is

->

&s

It is smooth by lemma 29.

jk: Is

Similarly

i(y)

be the space of

is the composition of the diagonal map

and a bilinear map.

j:

j 2:

= g iJX

is smooth.

E , let 92 S(E)

For any bundle

inner products on

i

)

Since matrix

inverse is a smooth map, by example 48, we know that is

Let

9?

is smooth.

s(SkT*)

(T*)k

defined by

to a product on

SkT*

gives a continuous linear map H s2 k ) -> HS(S2(SkT ) )

Combining

s (T*)

jk

s( s->z

and SkT *

qk

we define

S2 T* ) ,

j(g)

being the

k=O inner product on product are.

g

on

Z skT*C k=O T*(M) .

j

S2 T

induced by the inner

is smooth because

k

and

jk

__

___

__

-85-

I

Let Let

r:•

inclusion of inclusion

Hs HS

e) e)

-> S2T * ) _> SkT* s( Z SkT"C9 S2Ti~) S2T~) SkT"Cg c k=O k=O

s SkTC* Hs (S 22 (I: SkT+~3 HS(S~ k=O k~0

inclusion inclusion

continuous is aa continuous is

rr

maps. maps,

C cO

maps into into

the be the be

PR PR

2

s

2 kT S cc CO( SkTI1~S C0(,2 (C k=O

2

S T*)) T S2T,)n)

SS 2TT T

)

so it it is is smooth. smooth.

r o j 2 = roSor I~L2

f) f)

yg

fpg '

defined by

which is

O(M, IFt ) , f t CO(M,R ,) , feC

,

deterbeing deterbeing

pg M, pg M,

the correspondence yo . Then Then the correspondence

by the Riemannian metric mined mined by the Riemannian metric p~ f I"f,

on on

pO Pg

fix a a smooth measure We fix

3) 3)

q.e.d. q.e.d.

is smooth. is smooth.

2 C12

so

o i

which is defined by

p = fp

a continuous continuous associates with each continuous measure a, function on

.

M M

(M,2. ~R, C0 (M, -> cO Vs -> P3: ~ZS 1"3'

Let

Y ,

the measure associated to

P3

ProPosition Proposition 81:

Let

Prooof= Prooof:

from

SH

W fibre ~R, ribre

.

: Hs(SH) -> Hs(M,~

smooth. smooth. Then

Let

3

Therefore

(i.e.,

e(f)=ee f elf)

) .

Let Let

e(f)(p) e(f)(p)

e O o '13o p3 ooEE

map aa linear linear map M with M with

map continuous iilinear near map induces aa continuous

C-4 1W )) I~ HS(M, i: HS i: (M, IR

P3 = ioeo CL3

This is This is

bundle over X over trivial M ,, the trivial x M

i13:HS(SH)-'HS(M'IR)

by by

is

by by

= log( log(Jdet(yo* O exp a)/det(y 0O)).)·

.

is

P

smooth. is smooth, is

I" xX MM -> IR SH -> 13: SN Ct3:

~R

p

if

( 7 ) PO p = P3 P ~3(Y)Po

(a) IIS~(a)

to

SO that

=~Jd~t7~y~;JT/^Sd~-t~T~Y~n';T-~:~-;;O i j

C13(Y)

be defined

so that if

3(7) =4det(tyij)/det(TO )ij

locally

locally

))

e: H(M,

)

-> H(M,IR

e:HS(MP)--)HS(MVL)

= T

f(p)i/il) .

i=0 i=0

I )) CO(M, IR cO(M,

where

ee

)

is is

inclusion. be the inclusion.

E: ~)"f2S s -> E:?' ->

HS (SH) Hs(SH)

is is

-86-

·

tre aditeomorpnism terinea in

1).

--~T

hence

P3

L

is smootn.

and the extension is a smooth map.

on O~s

5: L(JS(S2T ) , Z

Let

k=0

SkT @ S2T

) X PR X CO(M,IR.)

B(Hs(S2T )) be the obvious map; that is,

->

· · I-

--

p: ' -> BO(H(S 2T )) extends to a map

Proposition 82:

Proof:



is the bilinear form induced on form on

H0((S 2T))

h

from the bilinear

which is defined by the measure

the inner product on homomorphism

HS(S2T * )

jý(h,g,f)

fpO '

J (S2T ) which comes from the

and the inner product

g

on

S Z

k_ k5 S T

2* 9 S-T

.

k=0 is clearly a continuous tri-linear map.

Io map.

8 B0(Hs(S2T )) is therefore a smooth ->

(P1,' 2,' 3 ):

It has been constructed so that

io

(Pl X P2 X

3)

The proposition follows. P: 0ýp s -> BO(Hs(S2 T*))

To insure that

Riemannian metric for 0 image of

s

defines a

we need only check that the

. consists solely of positive definite inner

products which induce the given topology on

Hs(S2T

)

.

We note the following: Lemma 83:

Let

E

be a vector bundle.

Then a norm on

H (E) is defined by a continuous inner product on and a positive continuous measure on on

i

E

M . The topology

H (E) is independent of the particular inner product

=

-87-

and measure which are chosen.

See 17, p. 148 (Palais states this lemma for smooth

Proof:

inner product and measure, but in his proof uses only the fact that they are continuous.) It is clear that the images of

p3

and

P2

consist

of continuous positive definite inner products and continuous positive measures, so together they induce the usual topology on

S2T)

is

is

s~Sk HO( k=0 a Riemannian metric, it

Hence to show the

1

enough to show:

0( Z skT* -> L(Hs(S 2 T*), Hi k=0

p: f

Proposition 84:

.

S2 T*) )

has image consisting of maps which are homeomorphisms into. is injective for all

P1 (Y)

Proof: for some

g

Hs (S2 T *

)

But this component is just 1l(Y)

Since (gn)

a sequence in

gn ->

0 .

da

'C

S T

then the com2T*

S T

2T* = S T

g

HS(S 2 T)

,

Pl()(gn) -> 0

implies

and To show this we will use coordinates

=

. SA (gn)

k=O

following statements:

---

sO(*

for, if

is injective, we need only show that for

examine the local situation. l(gn)

in 0?s

L(y)(g) = 0

,

which lies in

pl(y)(g)

ponent of is zero.

in

y

.

We note that if

7 = 70 * (exp a),

For the moment, we assume the

-88-

1)

g e HS (S 2 T * ) , k < s .

Let

Ak (g) =

D(g

)

P

+

terms of form

fD g

functions and

Iai

< k ; and

an

Hs-k

f'Dagi

f'

,

f5

k+l

gn

- >

0

implies

If

H[ s /

4 ]

in

Hk -

( S 2 T*

gn -> 0

We use ()(g)

f

f is a

Pk

is a sum of

[s/2] +1[s/2] Csk

Qk

2)

-> 0

k-1

-

is a sum of terms

function and

)

and

H k (S 2 T * ) .

means that for all

Note that

k < s ,

SAk(g)

k = 0

so we know that

We use induction on

is obvious.

large enough so that Ak+l k= l A (g) a a a

k a (g)

=

Ial=k D

->

0

. gn

-

> 0

Now the proposition follows by induction using 1):

S2 T )

HO(Sk T *

SkAag n -> 0O in

to prove the proposition.

SOa(gn) = gn

case

odd

C CO , then

in

in HO(SkT *® S2T)

Proof of

ven

k e

k/2

k odd

2

2)

where

k even

0

in

Qk(gn) -> 0 .

If

Da(gij)n ->

in

Da(gij)

-> 0

Qk(gn) -> 0 implies

q.e.d.

0

H[s / 4 ]

and

in this case also.

Z

D (gij )

in

Hk - 1

aj =k gn ->

in

0

Hk -1

-> 0.

implies

-90-

V.

The Group Action.

well.

The lemma follows.

-91-

Ow:.) :

s+d X

HS(32T

s+

H (S2T )

A(q,g) = q*(g) a

s+l

to a map H2(S2 T)

A

We can extend /

> Hs

2T )

by defining

In this case if we fix

as above.

*: HS(S2T * ) -> H(S2T ) , we get

and consider

the following:

q*: HS(S2T * ) -> HS(S2T )

Proposition 87:

is a continuous

linear map.

Proof:

Linearity is obvious.

I

II Ill , ,

113

-1(U

i )

II

corresponding

-1 (wi)

and

II 113 ,

1 ,

HS(S 2 T*)

be norms for

Ui , Wi ,

to sets

Let

as in Lemma 46.

II 1lIand II 13,

Looking at Tr in norms

it islocally

the sequence.

(1)

(u)) -> Hs, -1(

where (1)

is

P:

y

(3)

(2)

H (U1 ) -

piy ->

o

(PIy)

rj-1(Wi)

.

(1

is a map of the form it is continuous. tinuous.

(3)

Now fix _> ?s

y

by

is

a

which is continuous,

(2)

is norm-decreasing so it is con-

II 113, *?

t2ý

as described in corollary 31, so

This shows that

II II1 norm to the

s+ s:

is pe

,

-1))

and 3 is restriction to

o 4 -> (piY o 9)( I)(x),

the set

Hs

Hs (r-1(Ui)) ->

, ?((9)

r*

is continuous from the

norm. The proposition follows. (i.e.

y smooth) and define

= A(r,y) .

'li

-92-

4

Proposition 88: Proof:

is smooth.

sej s+l , and smooth exponential map

Fix

e: T(M) -> M .

We look at

i.e., consider

: T -> Z

V -> ?(V)

,

qi

aerined dI near zero in

T

V -> (y o e o V)(Te o V * TV)(Te o V * TV)

is

"o" means

where

4 in terms of a chart at

composition of maps and

means

"."

multiplication of matrices or linear maps. y

Since

by example 48.

Therefore

4 is smooth.

_, 89: Corollary

For all

Also

is smooth.

V -> Te o V

Hs+ 1 (U)

q.e.d.

,

yef

is continuous.

Everything remains smooth as above except

V -> y o e o V .

But since

we can use the weak ascertain that

is a diffeomorphism,

e o V

c-lemma 34, as in corollary 55, to

V ->

is continuous.

y o e o V

is continuous.

I

is the isotropy group of the action manifold structure to the coset space Definition 90:

I

Therefore

q.e.d.

Now we discuss the subgroup

---

is smooth

All multiplications are smooth by lemma 29.

HS(U) .

Proof:

Similarly

y o e o V

is clearly a continuous linear map from

V -> TV to

V ->

are smooth,

e

and

= (

E

S+1 A(,y)

of

Z

s+1

which

A , and give a s+1/I• = y

.

-93-

I

From the fact that

A

Z

St

is a subgroup of

consists only of r

C1

is a

means that

Cl

is an action it is clear that

.

Also we know that X

homeomorphisms.

homeomorphism such that

"tq E Iy ,

Therefore if q*(7) = 7 .

This

is an isometry with respect to the metric

r

any isometry of

7 . Clearly,

7 .

the group of isometries of

is

y

in

IT , so

is

I

The following results are

well known: Theorem 91:

Let

be a smooth compact manifold with

M

7 .

smooth Riemannian metric If

gn ->

g

Tgn ->

and

(i.e.,

If

Cm).

convergence

Iy

(1 < k

,

uniformly on

Tg

g

,

k .

A(g,p) = g(p)

IY

A: IY X M -> M

Also,

M

morphisms lie in

I

i(X)p = T(Id,p)A(X,O)

of



between the Lie algebra

Il

By

is

defined by

induces a natural identification,

vector fields on

C

is a compact Lie group.

Iy

the first part of this theorem, the topology of independent of

C1

M (i.e.,

is given the topology of uniform m)

.

I

be in

uniformly in all derivatives

gn -> g

convergence), then

(g n

Let

i ,

and the set O,

of

whose one parameter groups of diffeoi

is defined by

where

Id

is the identity of

IY

(see 12, 18). We shall now show that the inclusion a smooth map.

--- d

I

c•

s

is

We shall use the following form of the existence

7ON

-94-

theorem for differential equations: Lemma 92: zero, and

Let

Y

J

yO E

,

Y

uO EU ,

and let

Then there are subsets

be a smooth map.

map

R

containing

U be open sets of some Banach spaces

and

F , where

J,Y,U

be an open interval of

0, yO, uo

containing

E

f: J x Y X U -> F

Jo' Y0 , UO

of

such that there is a smooth

h: JO X Y0 x U0 -> U , and

D(t,y,u)h(l,0,0) = f(t,y,h(t,y,u)) , for all

(t,y,u)

in

JO X Y0 X U0

s

i: I T C j

Proposition 93:

is smooth. Id ,

We need only show smoothness at the identity

Proof:

r E I

at any

for,

94.

See 13, P.

Proof:

i = R -1 o i oR

,

T

,

and we know

that multiplication on the right is smooth for OV s

and

I . Id ,

To show smoothness at in

I

) s.

and

Since

that the exponential map from a neighborhood of

Id

in

I

.

-- L-

Z

Z ) of zero in

-> I

of zero in T (exp, Y) s

Id

is a Lie group, we know

exp: %

We let

As usual, a chart for (call it

Y

I

we look at charts at

is a diffeomorphism to a neighborhood

give a chart for

I

consists of a neighborhood

TId . As a map from

Y

to

Z

.

-95-

i

has the following form:

an element of M X M

this situation in local coordinates. coordinate neighborhood at

4:

and let of

T(M)

T(U) -> U x J.n

f(t,x,q) = v 2 o * o X(q) 7(9z1)

lemma

it is clear that are defined. JO

h

has range in

where

h

Assume

Y

small

defined by U .

Then the X

is locally

is defined by the above

U

and since

f(t,3X,q) = Af(t,X,q)

h(t,X,q) = h(l,tX,q)

is big enough to include h: Y0 -> HS(UO,U)

by

when both sides Y0

,

we can insure

1 h(X)(u) = h(l,X,u)

is smooth, all maps in the range of

and therefore

Hs .

(1)

following:

E),

be the local trivialization

Therefore, by shrinking

Define

since

a normal

being the zero vector field).

(yo

h: JO x Y0 X U0 ->

that

U

which corresponds to

q -> h(l,X,q)

i(X)

M (with respect to

in

f: iR X Y X U -> R.n

diffeomorphism

M ,

If

We examine

Take

induced by the coordinates.

enough so that

the map

p

.

M .

on

Tt

p -> E-'(p, r1 (p))

as

,

It generates

is the fixed exponential on

is the vector field

X

Then

C2)

X -> i(X)

X -- > h(X) -- > (p ->

h

are

C

is locally the

U)

(p,h(X)(p))) ->

(p ->

E1-p,h(X)(p))

The smoothness of this sequence is a particular case of the following lemma:

1

-96-

Lemma 94:

Let

A be a linear subspace of

is a Banach space, and let

1(x)

= 91 , where

(qt)

$: A -> V

Proof: *j

X .

Uo, W i

on

M

Ui -> Ui

'

(h, JO 0

Y0

be a neighborhood of zero in Y0

(one for each

for each that

i ,

(1),

(2),

max usU

PId o (3)

]j Du(7

4:

A

Ui

such that

'

as above).

2

Let

such that for each of the X E B , then

o ?j o X) II< E < 1

B -> TId

B

about

and such Id).

is locally the sequence

above.

usual norms on each Then assuming

4-1 o Id

4

is continuous in the

Hs W) Id 'o 4 is

k-l

we shall show that

Gateaux derivative converges locally and that it

defines the following map from

(+)

is small

Ui

(the chart of Z s

We shall show that

kth

is smooth

and trivializations

B c Yo , such that if

i)

4(B) c ?Id(GId) Then

Then

However, we insist that

enough so that there is another chart h: JO X YX0

be defined by

A

We pick charts

as usual.

s+l

which

is the one parameter family of

diffeomorphisms corresponding to near zero in

Cr(T(M))

dk (,ld o $)(b,v

1

B

to

Lk(A,TId)

... vk)(p) = Dk I d o $(b))(vl(p),... vk(p))

I

I

_1

·___

`

--l

`

-97-9-i

] c A, er (vii ,[v C -Idb

where

so

Dk(id o $(b))

near

p

and

B. B.

M .@Id

o •(b): (b): M M -> -> T(M) T(M)

,,

is defined using local coordinates (b)(p) .

?ld o

this map has range in derivatives of

p P

dk

bounded by those of

Since

o $(b)

is smooth, smootId

Lk(A,TId) . That is, the first

lld o 4)(bjv 1 ...

(vi]

s

are uniformly

vl)

lI o $(b)

and of

That the map is continuous will follow from the -1

proof of the continuity of We work locally on Then on

HS(Wi)

il

Ui

.

o

o ie which we now give.

and identify

we get the map

Y0 -> Hs(Ui'UI) -> Hs(Wi)

o~wihw

X -> h(X)

X

with Wi

v2 o

oX OX

from

By the existence theorem of

differential equations, (*)

h(l,X,p) = p +

We claim that if

eC

jO

Xh(t,x,p)dt

X n -> X ,

and therefore in

Hs (Ui,Ui)

Ih(1,X,P) h(l(l,X,p)l
h(Xn)

Therefore

)h= Dk(Id) th k

means the


h(X)

3(1,X ,p) (X

IDk D3 (1,

h(X n ) ->

h(X)

This tells us that Therefore Dk

-1 DId

B ->

X,p)

Id

IP oo

in

TId)



where

Ck , as in the

is continuous. o 4(b))

C

h(X)

and hence o nb->

n

Also i for all

o in

k , so the map

defined above is continuous.

...

vk ) ,

0

case.

Now it remains to show the existence of dk((Id o $)(b,v 1

,Xr? p)hl

p

h(Xn) ->

o 4(bn)) -> D(?

Lk(A,

Dh 3

o h)I

n

+

Therefore

h)

.

3

Therefore if we assume o h)

o

= DX ' Dkh + terms with

D3(1,x,p)(X o

3(l'VXJP)(X

3(13xpp)

Dk(lxp)(X

0

derivative in the third component

(i.e., D 3 h(v) = Dh(O,0,v))

Dk(,Xp)

+ f

and the equality at

(+) .

B

-99-

We note that

- h(1,X,p)

lim P(h(l,X+rV,p) r r -> 0

d(h)(X,V)(p)

By (*) d(h)(X,V)(p)

p

lim -

D3(1,X,p)h(V)

fI 1h(tx 0 r(Xh(t,x+rV,p)

- Xh(t,x,p) + rVh(t,x+rV,p))dt

At each

p , this converges to

J

1

DX

D2 (t,x,p)h(V) + Vh(t,x,p)dt

We will show that the convergence is uniform in in all derivatives. so

Vh(t,X+rV,p) ->

1(h(1,X+rV,p)

in

Vh(t,x,p)

- h(l,X,p))

p .

- Xh(t,x,p) )

- DX * D2(tx,p)h(V)dt + sr , r 2(t,x,p) Er ->

IIDX 11011 .O(h(t,X+rV,p) Now

Hence

- D 2 (1,x,p)h(V)

1

r Xh(t,x+rV,p) f01(X

=

h(X+rV) -> h(X)

We know from above that

- h(t,X,p))

CO convergence follows since

0

- D2 (t,x,p)h(V) I0

II DX 110

r



< E< 1

convergence follows as before, Just differentiate everything with respect to Assuming

p .

dk-l(h)(X,Vl ..* vk-l)

by the same method and

(+)

holds.

we get

dk(h)(X,V 1

..

Vk)

This proves the lemma

i

-100-

and also shows that

i: I

Definition 95:

X

Let

sional manifolds,

P & X ,

f

and

Y

f: X -> Y

is smooth.

be smooth infinite dimen-

a smooth map.

Tpf: Tp(X) -> Tf(p)(Y)

If at each

Tpf(Tp(X))

is injective,

Tf(p)(Y) , and has a closed linear complement,

is closed in then

-> 1) s

injective and is a homeomorphism into, then f(x)

an embedding and

f is

If, in addition,

is called an immersion.

is called

f

is called a submanifold of

Y

(see 13,pp. 19-21). Remark:

If

Y

is a manifold based on a Hilbert space,

any closed linear subspace of plement, so

Tpf(Tp(X))

f

is an immersion if

is closed in

Proposition 96: Proof:

so

i

i

has a closed com-

Tf(p)(Y)

is injective and

Tpf

Tf(p)(Y) .

i: I -> $

s is an embedding.

is a smooth injective map and

Iy

is compact,

must be a homeomorphism onto a closed subset of

Therefore, we must show that Tp i(T p(I )) is closed in identity in

I

.

Let

Xe

Tpi Ti(p)(

is injective and that s)

.

, and let

We work at the 7t

be a one

parameter family of diffeomorphisms generated by t -> qt

and

is a smooth homomorphism

Ti(X) = T0 h(d

)

.

$

h: IR. ->

Z

X s

Therefore, if Ti(X) = 0 ,

s

-101-

Th()

0.

Ttoh(~) t0 dt

But then for any

= T(tt0) o Ti(X) = 0 .

map, so for all Hence

Ti

,

Hence

qt =Id .

h

o TIdi o TR -1

)

T (I

so

is a constant

Therefore

is injective at the identity.

T i = TR Also

,

ts

tO e

X = 0 .

For any

se1 injective everywhere.

Tni

is finite dimensional.

Therefore

Tpi(Tp(I )) is finite dimensional and hence closed in

qe.d.

Ti(p)(·b s) Remark: identify

As a result of the above proposition, we can I

i(I ) .

and

Proposition 97:

Let

c: I X

composition of mappings.

Proof:

Then

TId ,

E-l(p,ef eg(p))

D(f,g)c(uv)

= D

,efeg(p

(f,g)

Note that

f

) )E -

1

be defined as

is smooth.

c

takes

c .

s

Id X Id E 1I x)

We show smoothness at

a single chart p ->

)s_

.

Using

into the map is smooth, so

( 0 , Df e g ( p ) e ( u e g (p )

+ Deg(p)f • Dg(p)e (vg(p))) the right side of the above equation being continuous in the local norms as usual.

Since

u

is smooth, the exist-

ence of higher derivatives follows in the same way.

Now

,

-102-

pick

q EI

c = L

o1 R

fact that

,

s 2

o c o (L -1,

c

Then since

S.

R -1):

is smooth at

Iý X

Id X Id

->

,

and the

~

s ,

L

and

are smooth (propositions 53 and 54) implies that (9,C)

smooth at

Corollary 98: Then

S

Proof:

.

Let

the

c

R is

q.e.d.

S =

U

s TIdR (TId(I )) = TIdR

is a smooth subbundle of

T(V

It is clear that the fibres of

q( ) .

s)

S

are closed

(actually finite dimensional) linear subspaces of

T(Z s)

Therefore, we need only construct smooth local trivializations.

Let

U

be a chart about the identity

The chart map gives a trivialization which defines an identification of Let

a: U X X

-> T(U)

a(u,X) = T(Id,u) c(X,0) , ? o a: U x

,TId( Z

s))

C(u)(X,Y)

$(Id)

= 4(u)(X) + Y ,

?: T(U) Z U X TId( TU(Z s)

and

sZ S)

Z

TId(

s) , uU .

is smooth.

and it gives a smooth map

defined by

: U -> L(TId(

in

a(U,X) = TId Ru(X) ,

a

s)

C be a complement to @ to a map

and

so

-> U X TId(c

: U -> L(( Let

by

Id

in s3),

X

is the identity map on

$(u)(X) = 4, o a(u,X)

TId(Z s) TId(t s)) ,

and extend by

Y 6 C .

TId(,

s) .

is smooth

Therefore

-103-

by shrinking

U , we can insure that the range of

contains only linear isomorphisms.

Let

by P,'(v) = (F(u))-1 o *(V)

T(U)

V

T(U)

S(U) If

and

u = Y(V) .

diagram:

> UX TId(Z s)

...

> UX%

any point of O

is

Tj

i': T(U) -> U X TId(• S)

= (TIdRu)-1(V) ('(V) VE

If V E S , say V s TIdRu( O ) , Therefore we get the commutative

'

s+1

, applying

TId(RQ)

we get

a commutative diagram:

T(N (U)) -- > RN(U) X T (

l

S)

RUl

S(R (U)) ---

R (U) XT

c

)

This diagram gives us a smooth local trivialization of fore

S

induced from a trivialization of S

is a smooth subbundle.

T(b s) . There-

q.e.d.

Remark:

T(• s) has a global trivialization

T(

>

s)

.X

TId

s) defined by

V -> T (R )-1V) .

However, we do not know that this triivialization is smooth.

-104-

On the other hand, the proof of the above corollary shows that

S

because if

XEEC

X,rj -> TIdR

Lemma 99: and

Y

V,

I(X)

S

%

,

defined by

is a smooth map.

is an involutive subbundle; that is, if

are vector fields that lie in

elements of 0

X,Y

Won

Z = [V,W]

s

by

V

S , so does

X

[X,Y].

S

and define vector fields

= TIdR (X),

W = TIdRI(Y)

Z'i = TIdR

is the vector field

vector fields of form fore

xZ s -> S

We use the usual argument for Lie algebras ---

Proof: Take

does have a smooth global trivialization

V

(

Clearly

l[X,Y]) ..

But

S .

There-

span the fibres of

is involutive.

.

q.e.d.

Now we are ready to use the Frobenius theorem to

Z

construct the manifold

Theorem 100:

Let

subbundle of

T(X) .

neighborhood

W

X

of

S/I 7

be any manifold, Then for any

z

S

an involutive

z E X , there is a

and a diffeomorphism

ý: U X V -> W

(U,V

open neighborhoods of zero in Banach spaces) such

that

$(0,0) = z , the composition

T1 (U X V)C-- T(U X V) T onto

S(W) , where

whose fibres are

>

T(W)

T1 (U X V)

is a bundle isomorphism is the subbundle of

T(U X V)

Tx(U) X 0 C Tx(U) X Ty(V) = T(x,y)(U,V ) x E U, yE V.

::

·

-105-

See 13, pp. 91-96.

Proof:

The elements of vectors in

T(,Z s)

S

are (from their definition) the

which are tangent to some coset

that is the elements of

T(I

comes from the fact that

I

q) _C T(Z s) , R~(I )

I=

is

I

(the inclusion a submanifold).

Therefore, applying the above theorem to

S

we get

the following: Proposition 101: hood

W

of

For any

nr

z

s , there is

Iq and a diffeomorphism

that for any fixed

$: U X V -> W

v e V , $ r U X (v)

phism onto a neighborhood of

$(O,v)

a neighborsuch

is a diffeomor-

in the coset

I $(0,v) 7: D

Let give

J

s/IT

s

Z

s/I

Pick

be the usual projection, and

the quotient topology, so that

tinuous and open. for

s/Iy

near the identity coset. W a neighborhood of

U, V

$(U,o)

and

Lemma 102:

.

Id

so that

Since

a homeomorphism into, we can make

nW=

is con-

We shall define a differential structure

as in the above proposition.

IT

v

$: U X V -> W

i: Iy c> Z W

s

is

small enough so that

Clearly we can restrict

W so that

W are connected. There exists connected neighborhoods

U O, V0

of

-106-

zero, included in

Proof:

Clearly

Since L5 s W1,

hoods

4(uo,v) c ITI N WO

X V1 ) , where x(Ul

about zero included in UO, V0

Pick

E IT

n WOW

Let

a

al

4(2)

Then I to

f W1 a

CI

n W1 .

be a line in

a

(I

W0 .

in

respectively. (U

(u',v)

4,

c IT

0

from

Then

e

nn W c $(U

X V) .

C

0

Therefore, if

= v ,

IT

n

so

WO

a, ,

r E $(U,v

I7T NW0 c

in from

r

to

C

at

rq= C(u,v 0 ) 0

) .

Since

W(U 0 Ov)

.

q.e.d.

w o $ r (0 X V0 ): V0 ->> s/I

is a homeomorphism onto a neighborhood of Give J~ s/Iy

)

From the

we know that the tangent .

a

I

an-

Id to

.

(U1,0)

(-1(a-1

to

is a smooth curve

Because of this lemma,

Definition 103:

= W0 C W2

a -

Therefore

U1

T$(T(U)) , v

X V)

ITl n W0 .

be in

O(h)r

W)

was any element of

Remark:

V

is a smooth curve from

any point lies in a =

and

a

construction of

and

U

and let

Therefore in

r

are connected balls

U1 , V1

small enough so that

E WO ,

Pick

for some

is a topological group, we can pick neighbor2 W2-1 c W of Id such that W1 C W , 2

W 1 =

and

such that

1(UO,V)

WO =

in

WO = C(UO X VO)

v e V0

r a WO , there exists a

for any IY

U, V , such that if

I

in

) s/I

a manifold structure by

-107-

declaring that for each element ?P

= VooRTo

: VO 0 ->

It is clear that a neighborhood of

s/I is

7'

I T

Z

of

's/

is a chart at

T

injective and

Ig

.

O)

cq(V covers

q , so we need only check that the charts 1

are smoothly compatible. (1) (2)

form:

V ->

4(o,v)

(5 wV2

1 ) -1l( I(O,V),q-T

° o

"

(where defined) has

1(3

)(0,V)( ->

->

where

2

projection on the second factor.

4(O,V)C7 10) -_> I

: U0 X V0 -> V 0

( 0, V)_I

is the

The first three steps

are clearly smooth; the last two comprise the map V -

2

s/I s

Remark:

-1~((0,V)Cr

o

-1 )

which is also smooth.

is a manifold.

R

acting on g

Proposition 104:

s/I

in the usual way is smooth. S -s>)

7:

The map

smooth local cross section at any coset Proof:

Let X 1 :

hood of 1 by map

X(

Therefore

(IyTI)

Corollary 105:

s/I 7

-> Z

= R, o = T , and

A map

7 o

f: L

is smooth if and only if particular

7

is smooth.

s

/Ii I

S/I

f o07:

.*

be defined in a neighbor-

o%)X • 1 ?(

admits a

isa smooth is the identity map.

-> N

(N

s -> N

any manifold)

is smooth.

In

-108-

Proof:

Assume

f = f o V o 7

f o r .

is smooth.

f o

and X

is smooth because, using charts

becomes

W 2 : U X V -> V .

is smooth.

q.e.d.

Hence if

Near a coset are smooth so R f

o $

and

is smooth

I

N f

p'q ,

is. T

r o f

--

-109-

-s+l> 7s

4y:

Now let

be defined as before by

= 7 ,

47(I1)

4,

(also called

1' = al (a E I )

4

Proof: $o

V

.

smooth. j

-1

Ie

$ 7: ) s+l/I7 ->Z s

l*(7)

$

If

.

*a*(7) = - (7)

(al)*(7) =

then

is smooth and injective.

4

Therefore by the Corollary 105,

(I 71)

= $(I C)

implies

and

I q = I7

.

an immersion.

g*(7) =

*(7)

so

is

To do so we must show that

s+1/Iy) -> T (p)(

s)

is injective and

We shall first study the map

and in particular

TIdId: TId(Z s+l)

We recall that HS+l(T(M))

TId(b s+l) and

is naturally identified

T (911 s )

is identified with

With the above identification,

TId : Hs+ (T(M)) -> Hs(S2T*)

4 ,

> Ty(%s)

HS (S2T* ) Lemma 107:

is

q.e.d.

has closed range.

with

,

makes sense.

We are interested in showing that

Tp : Tp(

Since

is smooth by Proposition 88 and clearly

4 =

(T) = r (7)

=

$(I 7)

so the definition of Lemma 106:

7

(also called 4)

defines a map

) by

7

onto an Orbit through

5 s+l

The Map of

VI.

is the differential

-110-

a

operator

derivative of x).

a(x) = Ox(7 ) , (the Lie

defined by

y

with respect to the vector field

(See Example 67).

Proof:

We know that

a

and

TIdI

are both

Hs+1(T(M))

continuous linear maps from

HS(S 2 T*)

to

Therefore it is enough to show that they agree on the dense subset V E C'(T(M))

Cm(T(M))

. Then

V

of

(rtj .

tO

is the map

+1

C:K ->

smooth curve in g s+l p

->

.

Pick

generates a one parameter

group of diffeomorphisms out that the map

HS(T(M))

We first point t ->

by

tis a

whose tangent at any point (p)

V

, an element of

o

Smoothness near

Id

.

T to

is shown in Lemma 94.

Smoothness

everywhere then follows from the fact that right

multiplication is smooth in 2

s and ~

.

From this it is also clear that the tangent to C at

C(t)

is

V o

Now to compute to the curve

d d

,

o c(t)) p))o

(tY

since

i

HS(S

2 T*)

is

a(V) .

Tit

e T't(

TIdP(V)

s+1) . we look at the tangent

o C(t) , at zero, =

.

d (())I

is by definition

is smooth,

~

(

At each point

so it must be the map

Therefore

@V(7)p

o C(t))

pEM ,

exists in

p ->

@V(7)p

which

The lemma follows.

We now look at

Ty

for any

'q

s

We

7

7 ...

*": HS(S 2 T*) -> HS(S 2 T *)

the continuous linear map =

A s*ts , Tq

so for any

Proposition 108:

Proof:

p =

T*

o

f

* : HS(S2T*) -> H s (S2 T )

(7 s+l))

T I(T

oT 1(R -,) (T (

•* o

is a restriction of

s

V

s>-V

i":

recall that

S+l))

o R

for

Ti o

o R

(C)

= T*(TI-1)*(7) = (*(7) = "T*(CT-1)

T 7T/= TT7"* o TId

o TTI (R

1 T1

T-

-1 )

T I(T (I

so

Therefore

s+l))

4 +1)) TI o a o T(R _l)(T (T Corollary 109:

T i(T I (

s+l))

and

Ti

1)

T I(R

are linear isomorphisms.

Our next goal is to show that closed in

TP(T)( ?)

'

TIdI (TId(b s+l))

HS( 2T )

are isomorphic subspaces of Proof:

and

) = HS (S 2 T*) .

T T(T (

s+l))

is

Because of the

above corollary it is sufficient to do this for the case

T = Id .

so we examine

We know

TIdi = a: H s+ l (T(M)) -> H(S

a . To do this we must use a few

properties of differential operators which we now explain. Let

E

and

F

be vector bundles on

D: C'(E) -> C (F) be a operator.

kth

Then as we know,

M

and let

order differential D

induces a linear

2T

) ,

r__ -112-

D: Jk(E) -> F .

bundle map:

0 ->SkT*

sequence:

Definition 110:

x E

k>

For any

k(E) -> Jk-1(E) -> 0

ot(D)(e) = D o ik(t x t ...

Let

t .

D at

and

so that

D

is called elliptic if t

T*(M)

in

p be a smooth measure for

and let

HO(E)

defined by

have smooth inner products

F

(')F

and

(')E M

E

p

t x e) , which is called

is bijective for all non-zero

ot(D)

t E T (M) ,

p E M and

ot(D): Ep -> Fp

there is a linear map

the symbol of

Also we have the exact

and

Ho(F)

have explicit inner

products. Definition 111:

A

D : C (F) -> CO(E) e

E

Cm(E)

is an adjoint of

f

E

See 17,

if

if for all

D

has a unique

D*

pp. 70-72.

fM (De,f) =

H-k (F)

D

(Def)Fdp = fM (e,D*f)Edp

Every operator

adjoint which we call

Note:

fM

, f E C (F) '

Proposition 112:

Proof:

order differential operator

kth

j

(e,D*f)

s > k .

for all

e

E

HS(E) ,

This is true because the

formula holds on the dense subset of smooth sections, and

D and

D

are continuous linear maps.

-113-

Proposition 113: at(D*) o at(D)

and

(ot(D))*: Fp -> Ep map

Proof:

Ep

at(D*) = (Gt(D))*

where

is the adjoint of the linear

at(D): Ep -> Fp

products on

t c T*(M) , ft(D* o D) =

For all

and

with respect to the inner Fp .

See 17, pp. 68 and 73.

Corollary 114:

If for all non-zero

ot(D): E -> Fp

t E T*(M) ,

is injective, then

D* o D: Cw(E) -> C (E) is an elliptic operator. Proof:

Any injective linear map followed by its adjoint

is an isomorphism. Since D* o D

Apply this fact to

fM (D* o De,e') = fM (De,De') = fM (e,D* o De')

is its own adjoint.

be the extension of

Let

If

D

is a kth

operator from

to

F

then

E

1) Ker D = Ker Ds CO(E)

Ds: Hs(E) -> Hs-k(F)

D: C"(E) -> Co(F) .

Proposition 115:

subspace of

at(D) .

order elliptic

is a finite dimensional

and similarly

a finite dimensional subspace of

Ker D* = Ker Ds Ce(F) .

2) Hs-k(F) = im Ds (DKer D* , in particular im Ds Proof:

is closed in H-k(F). See 17, 178-179.

,

-114-

Assume

a continuous linear map.

(algebraic) linear complement to B .

closed in

f: A -> B

be Banach spaces

A,B

Let

Lemma 116:

is an C

and

f(A)

is

B , and

closed in

f(A)

Then

CC B

B = f(A) ( C See 17, p.

Proof:

119.

D: Cc(E) -> Ca(F)

Now let

t

that for all non-zero Then

injective.

D* o D

have the property is

T*(M) , ot(D)

in

is a self-adjoint elliptic

operator. Proposition 117:

o D)s+k * Ker D

im (D and

s > k , Hs-k(E) =

For

o D.

Also

o D = Ker D

im (D* o D)s+k = im D The first statement follows from 2) of

Proof:

adjoint.

Let

or

HO(E) Clearly

(

)o

Ho(F) .

be the inner product on

Then

(e,e')o = fM (e,e') dp

Ker D* o D DKer D .

O = (D*De,e) 0 = (De,De) Ker D* o D = Ker D .

im (D* o D)s+k Cim Ds Ds+ k .

is its own

D* o D

Proposition 115, and the fact

D

Ker D

o

Also if De = 0 .

so

Furthermore

because

it

is

D*De = O , Hence clear that

(D* o D)s+k =

Therefore in view of the first statement

we need only show that

im D*

n Ker

soD=

D* o D = t0) 0}o

or

-115-

that

im D* (Ker S

then

D o D*f = 0

e = D*f = 0 .

De = 0

(D o D*f,f)

so

0

and

e = D*f

= 0 , or

im Ds+k

is closed in

HS(F) , and

s+k e Ker D s .

Since

Proof:

If

The proposition follows.

Corollary 118: HS(F) = im D

D = (0 .

Ker D*

is

s

closed we need only show

that the above direct sum is true in the algebraic sense. is

closed and the sum is

fe Hs(F)

topological. im Ds+k s+k

We first show that let

im Ds+k s+k

Then by Lemma 116 it follows that

D*f = 0

such that

Then

0 = (D* o D+ke,e)

f =

.

Now we show

Ds o Ds+k(H s+k(E))

and

+ Ker D s

, and

(0C

f = Ds+ke .

= (Ds+ke,D+ke)O

im D

HS(F) = D*-1(Ds(H s (F)) * s+k

Ker D*= s

so

spans

Hs(F)

Ds(HS(F))

by the above proposition.

Therefore

Hs(F) = Ds 1 (Dss o Dsk(Hs+k(E))) s+k = Ker D

+ Ds(H+ k

E))

The corollary follows. Now we apply the above corollary to the first order operator Proposition 119:

a . Let

as+l: Hs+1(T(M)) -> H(S2T*)

im as+1

Then

is closed and has closed complement in

) H s (S2T2

Proof:

t e Tp(M)

t / 0 ,

and

local coordinates of

p c M

All we need to show is that for any

7

p , and let

be such that

n xn

*

x

is injective.

Gt(a)

dx

Let

U

fE Ci(U,

)

V = vi

= t , f(p) = 0 , let

(df)

Pick

on a neighborhood

dx

if

8xI

p

be a smooth vector field defined near

p . Then

ot(C)(Vp)= fV(y)p i

fk

S(fv

k

6f

But f(p) = 0 , so gfV()p

dxi Q dxj

vk))

- +-

+ k .(f

f vk)

vk + ki(f 6

(Yki xj

6f

+ Ykj xi

k

p

j dxi 0 dx

From this formula we see that if of

T*(M)

induced by

corresponding to

implies

t = 0

is the element

under the isomorphism

fV(Y)p = t (V*

y ,

ot(a)(Vp) = 0

V

V*

+ V* or

0

t

.

Vp = 0 .

Therefore The

proposition follows. Now we know that for all

T I

is

closed in

is clearly onto

T,(,)( *

Tw( )(

s

E e Z s+1 , the image of ) .

= $o

and

s+/Iy) , so the image of

TT7

T V

-117-

TV

)

In order to show that

$

is the same as that of

also show that

T$

is an immersion, we must

is injective on each tangent space.

qie

Since at any

.

s+l , T•? = Tv())) o To

, it

is

enough to show the following: Proposition 120: then

a(V) = 0 ,

Hence if

a(V) = 0 V

T

(V)

r = Id

.

Here we

1

implies

is

= 0 ,

I7 .

Therefore

, so

T P(V) = 0 only if

RH o

I7

Tr(Id)RT

o TId

= 0

lies in the identity

For any

Therefore for

(V) = 0 .

->31•

is smooth, so o R -1 =

GV(7)

i E~s s+ V

T (S

s+l)

T R _ 1 (V) E Ker a , and this occurs

o TR

be defined by o v

v((,t))

TIdv(V) = 0

only if

T

But if

the one parameter group generated by

o*a o TR -1 .

=

TIdT(V) = O .

is a smooth vector field and

(It)

(qt )C

coset

R

and

s+1)

We first consider the case

need to show

TY

T (

V

T v(V) = 0 .

Proof:

V ,

If

R

.

R, o v = 7 oR

is smooth.

. Hence if

o TTR

R1 :)

Let

TId

s+/I / .

->

s+1/I

Therefore

Also

o T R _1 (V) = 0 ,

T v(V) = 0 .

_ 1 (V) = 0 , so

q.e.d.

We now have shown:

Proposition 121:

7: z s+1/I 7 ->q

s

is an injective

immersion.

1

VII.

(q-l)*

a diffeomorphism with inverse We shall show that

is

>

s

q*:

we know

s+l

11 Eo

For any

s

Acts Isometrically on

The Group ,Z s+1

is an isometry --- that it

i*

p: ft S -> B(Hs(S 2 T*))

preserves the Riemannian metric

This demonstration is a step by step examination of the p .

construction of

r

Assume

is smooth. T(M)

inner products on

7 , D

tively, and let

>


; i.e.,

by

' T-1W).

Dk, Dk

induced by

y(

V

XW) .

Therefore,

T7XW))

which is

D' o Trjno)

Also

= n*(Y)(Tr -

If

X ->

X -> Tr*(y( 7

X -> q*(Y)( VX'T

Corollary 124: on

w

W

.

Dc* Do is the map x ->

M and

1-form on

1

W) .

Hence

-1

W) .

The lemma follows.

is

are the covariant derivatives

y and

*(7y)

respectively, then

-120-

r*

o Dk = Dk o

Proof: D'

where

V*

7*

(T*)k

and

Follows directly from the definitions of

and the fact that if

T~* (t~o)

= TI*(-)

Let (T*)k

acts on


k

Let

(T*)

k

and

(T*)

a

,

D

TIi*(C(a)
k

r~*(Ty)

respectively.

=

'

be smooth sections of

g, h be the sections of

and




S((

(q*(g), q*(h))n

4

s+l

Since

)n

(

4'

(

T*(h))' g*(g),

nl

o q*(h))

n1

I )* (g ) -> g a•nd (-nl)*(h) -> h n(n

d x(p)

-123-

(g,h) = (r*(g),

Therefore

l*(h))'

q.e.d.

This finishes the demonstration that

P

isometry on V? s

is an

s?

is a connected manifold, so the Riemannian

metric on 72

s

s

defines a metric on f

(The distance between

way.

1*

p

and

the lengths of all curves from

p

q

is the minimum of q .)

to

induces the given topology on 92 S ,

in the usual

This metric

(see 19, p. 158), and

any (Riemannian) isometry preserves the metric because it preserves the length of curves. A: V s+l X

We have shown that

uous in each variable separately.

s ->

S

is contin-

We are now in a posi-

tion to show joint continuity. Proposition 130:

A

is jointly continuous in its two

arguments. Proof:

Let

Yn -> 7

and

Tn

->

r

in "f

s

and BO s+1

respectively. A(A(Tnyn) ,Yn)=

=Tn*(Yn )

continuous. Ps(

n'

7 )

->

ps (*(yn) so

A

If 0 . ,

"

ps

Gn (y) ->

Tn(y)) ->

since

0 . q.e.d.

1n

Therefore

since

s , then

is the metric on 9

Therefore,

is continuous.

i*(7) = A(tl,•)

is

an isometry,

gn (n)

->

*(t)

,

-124-

v

We shall eventually look at the normal bundle on Zs+ I7

induced by the immersion

$

effect of the exponential map of "i s

and at the

on this bundle.

First, however, we want to show that the exponential map

.V

commutes with the action of

sl.

Specifically we shall

show: Proposition 131: exp: T(J s)

->

f: O

If f

s

s -> As

is an isometry and

is the exponential map defined

s

by the Riemannian structure on V

(exp defined in a

neighborhood of the zero section), then for any if

exp V

is defined,

exp(Tf(V))

V s T(~ )

is defined and

f exp(V) = exp(Tf(V)) Before we prove this proposition we examine the definition of

exp . We use the construction in 13, pp. 109-111.

Definition 132:

Let

and

s)) -> T*(*? s)

7*: T(T*(

Co

w(V) = v*(V)(TW'(V)) ,

V s T(T*(9. s))

be the map induced by T(V9S)

n = [r *(dw)

T*(t s)

(which will be noted

p

a bundle isomorphism

on

on

.

: T(

P

s) ->

be the bundle projections.

The canonical one-form

structure

': T*(*

7: T( fS) ->fs,

is defined by .

The Riemannian

( , )) on

s) -> T*(

from forms on

Define the two-form

0

5

s)

'bZs .

Let rl

T*(9?? s ) on

T( t

gives

to forms s)

by

s

-125-

If

Let

f:

(Tf)*

forms on

be the induced map of forms

T(

Lemma 133:

Proof:

)-> T(9 )

s , Tf: T(%

S ->

to

T(97 S)

S) •

If

Pick

f

is an isometry

5 s))

X,Y s TV(T(

(Tf )* (n) and extend them locally

to vector fields which also will be called (Tf)*(n)(X,Y) = n(TTf(X), =

dc(T

1

X

and

Y .

TTf(Y))

o TTf(X),

T F' o TTf(Y))

(2)

(1)

= (T r' o TTf(X))(Wc(T r' o TTf(Y))) - (T F o TTf(Y))(c(T r

(3) - c(T rl (1)

o TTf([X,Y]))

= (T P

o TTf(X))((Tf(V), Tr' o T l' o TTf(Y)))

= (T P

o TTf(X))((Tf(V), Tf o T7(Y)))

= X((V, TF(Y))) Similarly (2) (3)

= Y((V, T7(X))) = (Tf(V),

T-'

oTT'

o TTf([X,Y]))

= (V,Tr([X,Y])) . 1(X,Y) = X((V,= Tv(Y))) - Y((V, T7X)) - (V,T7([X,Y])) by the same reasoning.

q.e.d.

o TTf(X)))

_~

___

--

-126-

Let K: T(es) ->1k by K(V) =½(V,V)

is a

T(.7 s

one form on Lemma 134:

dK

on

There exists a unique vector field X

T(V 8s) such that for any vector field

on

T(Vo

s ),

n(Z,X) = -dK(X) .

Proof:

See 13, pp. 109-110.

Lemma 135: Proof:

If f

(Tf)*(K) = K , so

By lemma 133, on

T(

is an isometry of

(Tf )* ()

s,

9Or

TTf(Z) = z

(Tf)*(-dK) = -dK .

= 9 , so for any vector field

S),

(f-1 also an isometry)

S(TTf(Z), X) = (Tf-l)* n(TTf(Z),X) =

(Z, TTf-1(x))

= -dK(TTf

-1

(X))

= (Tf-')*(-dK)(X) = (-dK)(X) Hence

TTf(Z) = Z

Definition 136:

by the above lemma.

Let

V e T(

S)

integral curve of the vector field

Let Z

OV be the

such that

Pv(O) = V .

__

r

-127-

If

PV(1)

is defined,

exp(V) = y o PV(1) .

Proposition 137:

exp

exp(V) Z

is defined and

is called the exponential spray.

is defined on a neighborhood

of the zero section of

T(W s)

and if

9

Exp: 9 -> 9Ps X

s

by Exp(V) = (w(V), exp(V)), Exp

is a local diffeomorphism

near any zero tangent vector in

T(ft S)

Proof:

See 13, pp. 72-73.

Proof of Proposition 131: of

exp

BTf(V)

Itis clear from the definition

that we need only show the following:

PV

and

have the same domain of definition and

Tf o PV(t) = PTf(V)(t)

for all

t

To show this we need only check that integral curve of

Z , and

statement is obvious.

Tf o PV

is an

Tf o PV(O) = Tf(V) .

The latter

Also

d(Tf o Bv(t))t 0 = TTf(•(v(t))Ito) TTf(ZP(t))

in the domain.

= ZTf o Pv(t)

= TTf(ZV(t))

and

by lemma 135.

The proposition follows. As a final preparatory step to proving the slice theorem, we look locally at the normal bundle

by

4:

s+1/I 7 ->

Definition 138:

Let

v

induced

S X, Y

be two manifolds,

f: X -> Y

_

__E_________________~__

-128-

an immersion and Then

7: T(Y) -> Y

be the tangent bundle.

f*(T(Y)) = ((x,v) c X X T(Y)lf(x) = r(v))

is a bundle over

X

with fibre

It is called the pull back of Clearly if

f

f*(T(Y))X T(Y)

Tf(x)(Y

f ,

by

)

(see 13, pp. 38-39).

is injective, there is a natural map

v(f): f*(T(Y)) -> T(Y) by (x,v) -> v , which isalso injective. is an injective immersion

so we can find

be a neighborhood of the identity coset in

$ r U is an embedding or

that

$(U)

s+/I/

U

to

7

such

is a submanifold

of ~7 s Definition 139: and

V

Let

v(U) = (V E T(fs)Ilr(V) =

is normal to the subspace

Tu $(Tu(s+5

$(u)

I))

c $(U) .

This

iscalled the normal bundle of $(U) in9 Proposition 140:

v(U)

is a vector bundle and

exp

1 v(U)

is a diffeomorphism from a neighborhood of the zero section

of

v(U)

Proof:

to a neighborhood of See 13,

pp. 45-46,

$(U)

73-74,

in

?s

and 104-105.

i-ipw-

____

__

__

___

-129-

The Slice Theorem.

VIII.

Theorem 141: S

of

t)fs

, ,

r

I

,

i*(S) = S

There is a neighborhood

such that if rjE V 3)

->

U

and so that

V in 9

exp rv(U)

$(U)

E-ball about zero in in the set on which

Proof of q*(s)

1):

Iy

F: U X S ->

F

s

is a homeo-

sN

is a diffeomorphism when restricted

v (Id)(U), exp

preceding proposition).

exp

rj

is a submanifold of W? s J

to the set of vectors of length < E .

Since

, then

o S+ýly

a neighborhood of the identity coset,

E > 0 , so that

and

in

, and

8s+l/I

morphism onto a neighborhood of Pick

4

F(u,s) = ( C (u))*(s), then

is defined by

Proof:

I

is a local cross section at

s+'

8+1

7:h4

V of

rj (S) n S /

and

X2 : U -> 8

If

the identity for

7(Id)

there exists a submanifold

such that:

1) For all 2)

ys

At each

r

r v(U) Let

so that

s ES

S be an open S

is included

is a diffeomorphism (see

S = exp(S)

S is an embedding,

Pick

Let

and let

= q* o exp(t) = exp o T(j*)(t)

S is a submanifold. s = exp(t) by proposition 131.

_

_

_I

-130-

Since

II T(*)(t)I1= II t II
B

is clearly bijec-

Therefore, since

is bijective.

F

F

is

(where

is the right action), and

F-1 (W)= (p o exp-1(W), A(, -1i Therefore,

B ,

K

F(u,s) = A( ?C(u),s)

s ->9s

1),

I

F(u,s) = exp o K(u, exp-l(s)) .

s+l x×

as in

is a vector over

be the bundle projection map.

continuous since

a vector

is

4

B = (V e v(U),

Let

E

Then

, and therefore,

Tr*(t)

is a diffeomorphism on

exp

A

j

and assume

II

is a homeomorphism onto

be defined by

Tq*(t)

II TT*(t) 11< E .

and

T EV

s' = exp t' . ,

$(U) 4

i*(7) = 7

Proof of

tive.

= exp t'

being injective), so

Hence

Pick

s = exp t ,

Let

4•II

T*(t ) e v(U)

that

(U) .

, an element of

4

(exp

.

= exp T(q*)(t)

T* exp(t)

over

V = v

Let

C

and

W E exp B , o

continuous,

-1I o p o exp-1(W),W)) so

F

is

a homeomorphism.

q.e.d.

LIII__

__I_

-131-

IX.

The Smooth Situation.

Since

A:0

s+1

s _>

large

s , we know that

also.

In particular,

if

#5 /1

s

A:

is continuous for all

X 21-> ft ->

P, :

W

is continuous

is continuous, and

4 : ) /17

is given the quotient topology,

- >

is a continuous injection. •

/I

onto a closed subset of

/

4):

Proposition 142:

->/I .

is a homeomorphism

The proof will be a sequence

of lemmas.

Lemma 143:

set

(pi9

(n)

of

in

Given any sequence

of points in (qIm)

compact set. Cn(Pi)

n(Pn )

For a single

, so

->

pi '

well.

(nn)

qi

1 "

7m(Pi)

is a sequence in a

satisfies the lemma.

Then refine

Cn

Tn

so that

so that

(Pi ]

If

Cn(Pi

Cn(Pj+l)

)

= [Pl "'. Pj+,l

-> q1

for

converges as

This refinement satisfies the lemma.

Lemma 144: in

i , there is some

Hence it has a convergent subsequence

assume there is a sequence i < J .

and finite

M . There is a subsequence

such that for all

M , such that

Proof:

([I _ C

Fix

such that

7, r' Trm(y)

E

->

V? and let 7'

.

([m)

be a sequence

Then for any finite

___

__i_

__

-132-

collection of vectors

(Vi9

in

Wi

[rn)

, such that for all

and a subsequence

TCn(Vi ) -> Wi Proof:

Let

vectors

Also, if Vi

0,

i ,

Wi + 0 .

K be the maximum of the lengths of the

Vi

with respect to

.

K = max[('(Vi, Vi))1/2 i .

T(M) , there exist vectc

Y(TTJmV I ,

m,

qm*(y) -> y' , for each

Since

y'(Vi,Vi) .

mVi) ->

ciently large

'' .

Y''(TV,

Therefore, for suffl

qmVi)1/2 < 2K .

Let

S2 K(M) = [V s T(M)I(7'(V,V))1/2 < 2K) . S2 K(M) is comps and for large m , T,,m(V ) E S,,(M) . Now we can construct a subsequence as in the previous lemma.

and

(Y)(vivi) -> 0 so

W = 0 then

which is impossible. Lemma 145:

Let

e: T(M) -> M be .

K, %, and

spray defined by let

Km, fm

and

Z

y on

qme = emTm

Zm

"

Tm

is an

We use the same proof

be the function , two-form, ani

T(M)

(see lemmas 133-135), and

be the corresponding objects for

ThenK m=(TTm)*(K),

7*(') •m

Then

We have proved this in the case that

Let

=

the exponential map of .m(Y)

isometry (see proposition 131). here:

y'(Vivi)

q.e.d.

em: T(M) -> M be that of Proof:

If Vi

am =(Tm)*(n) T

m

(2 )

and ad

Zm = TT =T

-133-

PV(t)

Therefore if V,

Thm o PV(t)

is an integral curve of

Proposition 146: y' .

Z

through

TT1qmZ

through

e': T(M) -> M be the exponential

Let

Then

em -> e'

on compact subsets of Proof:

an integral curve of

The proposition follows.

T1m(V) .

map of

is

Let

-> Y'

T(M)

K', n' and

and spray defined by

Z'

7'

-(Y) L', Km ->

,

uniformly in all derivatives

be the function, two form T(M) .

on

m -> Q'

Then, since and

Zm -> Z :

all convergences being uniform in all derivatives on compact subsets. If

and

, e(V) =

V e T(M)

(1,V)

=

jO

V +

7

o

(1l,V)

w: T(M) ->

, where

Z(tv)dt, the integral curve of

M

Z

Convergence for such a case is proven in lemma 94.

Proof of Proposition 142: exists a number respect to

Since

M

is compact, there

E > 0 , such that any ball in

7) of radius less than

Let (pi

]

of

K =

max

E' > 0

(with

E lies entirely in

some normal coordinate neighborhood of there exists and

M

M .

Also for

7'

with the same property.

(y'(V,V)/7(V,V)]

VE T(M)

M such that for each

i ,

.

Pick a finite set Ui

is a normal

_ ___~i~

-134-

(with respect to

7) coordinate neighborhood centered at

U Ui = M . i (ViJ) j T

Pick

(vi )

i

such that for each fixed

is an orthonormal basis (with respect to

,

y)

of

(M) . (Wi J

Find Ck(Pi

that

(WiJ)

)

C T(M)

->

q1

Tk(Vi J )

and

is a basis of

ial :

such

for all

-

i,j

.

Tq (M) by lemma 144. i1

q = e(z aiJvi

j

)

Ui

about

Z(ai )

where

Ji

Pi

If

< 6

>

(-

1,- In 4

4

mi Th

Consider a fixed

i

q E Ui

Therefore,

ji

k(qk) = Cke(Zai Vi j ) =

[TC (ViJ)

so

)

[(k

and a subsequence

j

e o Trk(ZaiJVi j ) ,

Ck(q

so

)

Tk(ViJ)

->

Wi

ek -> e'

Also

is a bounded set.

k

uniformly on bounded sets,

-> e'(zai5 WiJ)

C(q) = e'(Z aiJiWiJ)

Let

Then

-

Ck

> C

on

Ui .

M be defining it on each M

and

6 = K-1/2min(S,E')

Pi ' of radius less than

and

Ck(q

The two constructions of must coincide. since

= e'

Ck

-

Therefore

> C

o L o

on each

e-1

Ui .

Since

to be a map on

C

(Rk) are maps on

C(q) , it is clear that for

->

)

Extend

Pi

where

1(q) C Ui '

e

(from Ui

Ui

q

and

-

> C

on

M .

: T (M) -> M is Pi

I

Ui ,)

is well defined on k

0 U

M

and

On

Ui ,

e

Tp (M) 1

.

_IIIIIC

______

I

-135-

and

L: Tp(M) -> Tqi(M) j

L(Z aiJViJ) = 2 aiJi

Loe e - (Ui)

is the linear map defined by

.

Since

is contained in a neighborhood of zero of

radius < E' (with respect to

e

'

r Lo

y'(WiJwiJ) < K ,

-1

ePi(Ui )

7').

Therefore

is a diffeomorphism.

a diffeomorphism onto a neighborhood of To show that

C

a diffeomorphism,

C(M)

compact, so

is closed in

Now we show

C

be the metrics on

respectively. covering

[Ui )

is open in

induced by

with respect to Ui

is injective.

So we consider

and

pk(Ck(p), k Ck(P) ->

p'(C(p),

k(q))

C*(7)

orbits of o

Since so

in

> a

Ck(q) ->

> I , so

C(q))

diffeomorphism. *(7) ->

such that

C(p),

M .

C t Ui

But

Hence Let

M C

p, pk

is

is is onto. and

7 , Ck(y) , and

p'

Y'

A be the Lebesque number of the

there is some

Then

C(pi)

Since

M .

is injective.

M

Let

is

is a diffeomorphism we need only

check that it is one to one and onto.

C(M)

i

Ui

Hence

p,q s Ui

a homeomorphism into.

.

for all C(q)

.

k .

Ck -> C in .

are closed.

C

UU i

p(p,q) > ('()

But

->

y'

Therefore,

C(p) + C(q) .

Let

We know

p, q E M where

C*(7) = 7'

V

p(p,q) < L ,

p ; i.e., if

0E

Hence

C

is a

,

This shows that the

Now we show

imI(7) ->

i*(Y)

.

s: 0/I1 Then

->

.

__

-136-

,

/I

1

in

->

-1)*()

(Tim o

then

I

h /Iy

in

U

that

(qm i ) n U = 0 .

that k in

E

Ck

-

U .

and

(Timi)

Contradiction.

/I 7 •

7 , so there is

a

and a diffeomorphism

C

Hence if

rml(y) ->

such

k ,

Therefore for large

.

C E I

->

such

(Tim

and a subsequence

(Tmi)*()

[k) of

> C

the desired conclusion.

and that there is a neighbor-> 7ym(7)

hood

subsequence

T-> I~---

m

I

Therefore assume of

I q m o i-1 _> Iy

If we can show that

.

y

IyT m ->

IY

The proposition follows.

/I-> /:

Using the fact that

is a homeomors , one

phism into, and using the invariant metric on " can prove the following:

,

y s

If for

Theorem 147:

then there exists a neighborhood all

7' e V ,

any

y E

such that

(7 ('=

t'

I7

is trivial.

there is a I,

,/ III

= (Id) . = (Id))

7'

is the trivial group,

I V

7 , such that for

of

Also if

dim M > 1 , for

in any neighborhood of

y ,

That is, the set

is open and dense in

.

We call

it the set of generic metrics.

Preparatory to the proof of this theorem, we construct an invariant metric on / t

metric on 9r

.

For

t

s , let

pt

be the

induced by its Riemannian structure as

defined in section IV.

-137-

Let P('Y"Y )

->R

X*

p:t

1

=PteYorY

t>s

Pt

is a metric on Et which is invariant

p

under the action of Jb

topology on V

-

p is an invariant metric is immediate from

That

in

7y

*1

pt

t

p(yn,7

therefore,

and which induces the given

.

the fact that each n

"

)

1+Pt(YsT

2

Proposition 148:

Proof:

by

for all )

( 7n '

y) -> 0 .

t ,

implies that

n -> 7

n -> 0 Pt( YY)

implies that

Therefore,

->

Y,

7)

Conversely if

then for each

n

pt(y n ,

t , so

-> O . Pt

yn -> 7

is. Also, if

C1

in the

y ->

Ps ( n,'

But

7)

0 ,

-> 0

sense and generally y in the or

If

p(Y,7') < e

and

so

p(q*(y),y)

for all

-> 0 , and

p(Yn,7) ->

k ,

Ck

in 2

C

sense.

Yn -> 7,

This proves the proposition. Proof of Theorem 147: then

p(Tj*(Y),7')

choosing max

< E

< 2s .

T E

Is,

Hence by

a small neighborhood we can insure that

V

(p(T*(y),y))

is small.

Since

4

:.

->

9t

is a

homeomorphism into, this means that we can require that the set

U y'sV

(r}

is within any fixed neighborhood of

Id

-138-

.

in

Hence to prove the openness of

find a neighborhood of

Id

in

V

p e M

Fix a point

E V,

7'

and let

coordinate neighborhood of respectto

y , and a neighborhood

of

such that if

2

ýb we need only

p

W

W.

I,

be a convex normal

U

with radius

3a

(with

y). e r Tp(M)

That is,

is a diffeomorphism from the 3a

open ball about zero of radius points

ql, q2

ql

to

q2

q2

are within

onto

U , and for any

U , there is a unique geodesic from

in

6a .

of length less than e

some distance

Also, if

ql

and

p , so is the entire

of

geodesic segment between them, so this geodesic lies entirely inside

U ,

(for existence of

U

see

3,

pp, 246-248). Pick 7y' a V,

V

a neighborhood of

U

contains a convex neighborhood of

radius bigger than any

E IV,

where

d

and

y' Let

and

, ->

d'

such that for all

d(q,q(q)) < a

are the metrics on

with

y' , and for

with respect to

q EM ,

p

and

d'(q,q(q)) < a ,

M induced by

respectively. 0n

dim M = n . f: I

and

2a

y

On

be the orthogonal group for Given

7' E V ,

$pn

where

we shall define a map

7

7

We first

F(U)

X

define a function

: U -> F(U)

is the bundle of orthogonal frames over

where

U

with

respect to

7' .

over

p

and define

X. (q) to be the parallel translate of

F

to

q

q E Iy,

= FIFp

X(r((p))

Claim:

r s 1.,

If

of

where

Frj

the angle between

Case I:

= Id .

x

q

by the equation

is the map induced by

r

M

on the

f(Qi)(x)

rq(p) = p .

Assume

and

(n-l) sphere such that x

is greater than or

Then

f(rI)

= Id

0

n

only

But it is well known that if

.

TpB = Id: T (M) -> 1

to

7/2

Trj = Id

if

p

1 + Id , then there is a power

and

1 and a point

equal to

f(I)

and define

on the bundle of frames over

1i

Fp

along the unique shortest goedesic from

Now pick f(Tr)

Fix an orthogonal frame

T (M)

and

Ti

is an isometry, then

(See 12, pp. 131, 138,)

Clearly,

f(i) =

in this case, so our claim is an elementary property of

Case II:

n(p)

1p

Assume the claim false and let of

be some power

. Let

from

r

p

b = d'(C(p),p) to

C(p)

and

and let r(p)

to

g!, g 2 Tq(p)

be the geodesics respectively.

.

Let

V1

be the tangent to

tangent to and

V2

g2

at

C(p) .

p

of radius

b

geodesic Hence

V2 g2

Let

Sb

and

V2

the

is farther from

d' .

C(p) E Sb Sb

Sb , so part of the

p

than the distance

, because

are in a convex neighborhood of

V1

be the sphere

is perpendicular to

V1

points outside of

d'(p,9T(p)) > b

C(p)

with respect to

and by the Gauss lemma, Therefore

at

Then the angle between

v/2 .

is less than

around

gl

Tr(p)

and

b

C(p)

p .

Using induction we.can state the following: ),

The sequence farther from

2(, p),

13 (p),...

gets continually

p .

We recall a trivial property of compact groups: Lemma 149:

If

an element

1

G of

is a compact topological group, given G

and a neighborhood

identity, there exists a power

i

of

W q

of the

such that

i

Proof:

Assume the lemma false, and let

hood of the identity such that check that the set set cover

(1iV)

V

VV- 1 C W .

(qiV)

of

The lemma follows.

[ i]

Then we can

is a disjoint collection.

is closed and therefore compact.

(3i)

be a neighbor-

But the

has no finite subcover.

The

IV,

is a compact group and by the above, p , so

does not approach

7i(p) Id .

li

does not approach

This is a contradiction; it establishes our claim. We shall now

show that by shrinking 7'

we can insure that if

on the

E V ,

V

slightly,

, there is no

1 E I,

(n-l) sphere such that the angle between

f(7)(x)

7/2 .

is bigger than or equal to

x

x

and

This, of

course, will conclude our proof of the openness of b Assume that

a

is small enough such that for all

q EM

,

to a

3a -ball about zero, and restrict

any

e rTq(M)

is a diffeomorphism when restricted

'' E V , the above is

true for a

e' , the exponential associated to Then, for any

if v', v e Tq(M)) between

v

and

Y(v,v') < 0) restrict

v'

then

.

so that for

2a -ball using

7'

q E U , there is

and

V

a

c > 0

such that

I1v II = I v' JI= a and the angle

is not less than

7/2 ,(i.e.,

d(e(v), e(v')) > 40 .

V , so that for

7'

e V

We further

and corresponding

v,v'

d'(e'(v), e'(v')) > 3c • Now find

and

Vp E Tp(M)

and

vq

h, a > h > 0 , such that if

such that

v II

a , then if q E U

is the parallel translate of

the shortest geodesic through

p

d(p,q) < h

and

vp

to

q , then

q

along

7~ -142-

d(e(v), e(v q)) < c 7'

a V,

.

Restrict

q E M ,

and

Ny

vq

Vp

Pick

vq

c .

E I,

vp

< d'(e'(vp),

e'(v p)

h Si%± U

C

be the translate of

e'(T"v ) =

since

T M

a .

e'(vq))

to

e

th1- th1 e

CZ

IV,

ra

q = r(p) ,

and let

q . Then + d'(e'(vq),

so the first therm on the right

Also the second term is less than

e'(v q))

is less than

e'(vq)) < 2c

y' E V ,

e'(TI(vp)))

is a parallel translate of

d'(e'(Tv p),

d'(e'(v p),

< min(c,h)

,

is

q ,

so that for all

so that for any

VT

r

vp

is less than

vp

'

Fc%

d'(e'(Trqvp),

But

V

d'(q,r(q))

i ow

length of

let

v p,

and corresponding

Finally restrict

V

< 3c

7/2.

vp .

2c

Therefore,

so the angle between This shows that

TTIvp

and

is open.

Now to show that A& is dense we look at the curvature tensor and perturb it slightly. T_(M) f

be the non-zero vectors in

: T_(M) -> j-

Ric(X,X) f

Fix

by

f (X)

ty E

T(M) . Let

= Ric(X,X)/<X,X>

,

where

denotes the Ricci curvature, (see 11, p. 95).

is a smooth function on

T_(M)

and if

set of normal coordinates in a neighborhood such that the components matrix at

and let

p, then

gij

of

y

(Ixi U

are a of

p ,

are the identity

e'(vp))

2 gni

2gnn

Z xnx nix+ i

xi2

1 gii

f (X ) =

(

i nX

2

n

-

(This is a direct calculation using the standard formulae for curvature and Let to

S(M)

SE

S(M)

has a maximum on

- (0o ,

maximum of near

connections, see 11, p. 63).

be the set of unit vectors in f

').

affine

f

f

T (M).

y so that

fy,

i

that for all

i, j, k ,

7' ~

1

nates at

+ Ei p

j)

for

ij

(

) =

arbitrarily

U .

Then

2iini

i i sx

n

on

-

2

U

U , such

= 0 . y' = '

p ,

ir txnx

off of

+

axi 2

and

a

Eij 5

CO(U) , such that

(Ir n=l, =

U

2nn

..

f'7 ry'(Xn) > f yn (X ) . (

y'

f ,(Xn) = f (X )

It is clear that if n > 1 , there are rarily close to zero in

Let

are normal coordi-

(xl)

y' also, and at

+ 1

( ij)

whose support is in

M defined by: over

y'

has a greater maximum.

(symmetric in

be the metric on

and since for

We shall find

Take a set of smooth functions

and

(with respect

, this maximum is the

(mX) = f (X)

on

M

f7 =f 7

= O.)

This construction tells us the following:

arbit-

-144-

1)

At any point

p

of

M

and unit vector

X E T (M) , and for any neighborhood find a Riemannian metric

such that

7'

=

y

U

of

p , we can

y' arbitrarily near to

off of

U

and

7

fy,(X) > f(X) .

Also, the following is immediate from the definition of

f :

2)

y

For any

EWZ

, if

rl6

,1 f (TrX)

= f (X)

X E T_(M) .

for any

Furthermore, from the proof of the openness of A we know: 3) p

in

p s M , there are neighborhoods

For any

M , and

V

of

where

, there exists d

in 9f

y' e V ,

such that for any la I,

7

I

= (Id) (,

q a U

is the metric on

and there is

of

6 > 0

or for some d(q,7(q)) > 6

such that

M

a

U

induced by

y7

Using the above three statements, it is easy to show the denseness of maximal at of

p

&

X e Tp(M) . 3a

of radius b

Perturb

y

. We select Let

U

p EM

so that

f

be a convex neighborhood

for which

3)

holds.

Pick

b0 < a

b on

fy(X) > f (X) .

B

as in

1)

Then for any

is

to get Y

70

T(M) ,

so that such that

b

T_(Bp

Y

)

(Y)

f T(X) >f

,

.

b

S2 a = (g e Mld

Let metric on •I 1

(f

max

such that there is

(Y))

.

X1

&

and

such that

Tpl(M)

(Xl)

f

=

0

yeT (A1 )

and

q E S2 a

Then if

2a - b 0 < d 0 (p,q) < 2a + b )

EM ,

pl E A1

pick

70 .

is the

do

where

(p,q) = 2a)

< d 0 (p,j(q)) < 2a + b

2a - b

,

0

B 0

rT(p)

so

defined by

M

A1 = (q

Let

b 0)

E T_(B

, Tr1(X)

SI0

2) , if

Therefore by

0 b

f o ( XI ) < f1

(X )

( X)=

< fl

b

and

induced by

71 ,

f 0(X) .

e(Pl EB

i

.

A2 =

and let

Let q

a-b1 J d l (p,q) < a + bl).

X2 a Tp (M)

so as to maximize

Continue inductively to get

pl ...

pn

-. and

let

->

1

=

E

n

(pp

Ie I

dl

be the metric

M12a - bO J dl(p,q) < 2a,

Pick

P2 E A2

iL )

and

fyl b0

...

bn

A

I

A nn

= (Id)

be the exponential map of

M X M

,-,-1, Y

Then if

71 ... 'n . We claim that

and

En: T (M)

Let

pl

about

b

)

q(p) EB 0 p

and

B 1 11

bl-ball

such that

71

to get

on a

y0

Now perturb

,

i=1,...

Tn

bO

Case I:

(Yi)

for any

q

are independent vectors in

U ,

q = en(2 yiYi)

r E E1

so if

Thus

T (M) ,

i

small 71(q)

= e n(

yT 1 (Yi))

But by making

TQY

we can insure that

bn

bn0

is arbitrarily close to

i

Yi '

(since the inequalities

f n(x)n < f nn(X)

(Xn) < f n(Xn-1 ) "

(*) f

and (**)

fT(X)

> max

bi)f

YE T(Ai-B Pi) n

(YY) n

b

imply that that

means

r(q)

b

and 7(pi) E B '), and hence T(p) E B 0 p q . By 3)) , this is arbitrarily close to q . By this

q = Id .

Case II:

are not independent, move the

If (Yi)

slightly so that they are.

Clearly this can be done (*)

and

by the argument of case I,

Iy

without annulling the inequalities , Therefore,

Now it is clear that of theorem 147 is

t

is dense in •

complete.

p

(**) = (Id)

, so the proof

.

-147-

X.

Further Research.

The present work gives rise to several avenues of additional research.

Perhaps the broadest one is that of Hs

further applications of another.

maps from one manifold to

This is the first instance known to the author

of the use of

Hs

possibilities.

maps, but they should have many other

They are preferable to

in that the manifold of

Hs

Ck

functions

maps is based on a Hilbert

space rather than a Banach space, and in that they are a more natural tool when discussing the range and Kernel of a differential operator. One thing that might be done using them, is to study the orbits of XE

(or .0 s)

in spaces of other structures

(e.g. conformal, complex, semi-Riemannian) on a compact manifold.

Eells and Earle have done this with conformal dim M = 2 , (see

structures in the case

8).

As to more particular results, it would be nice to

show that the map

~: V

s+1/I/ _> %

onto a closed subset of V

s

is a homeomorphism

and by this, to get the

full classical slice theorem.

This would tell us that

the orbits of 0 s+l

are closed and give a first

in W

S

step towards investigation of the quotient space

7 s/iO

s+l

REFERENCES

1.

Abraham, R., Lectures of Smale on Differential Topology, Notes at Columbia University, 1962-63.

2.

Bers, L., John, F., and Schechter, M., Partial Differential Equations, Interscience, New York, 1964.

3.

Bishop, R., and Crittendon, R., Geometry of Manifolds, Academic Press, New York, 1964.

4.

Borel, A., Seminar on Transformation Groups, Princeton University Press, Princeton, N. J., 1960.

5.

Theory of Lie Groups, Princeton Chevalley, C., University Press, 1946.

5a.

Diendonne, J., Foundations of Modern Analysis, Academic Press, New York, 1961.

6. Eells, J. Jr., On the Geometry of Function Spaces, Symp. Inter. de Topologia Algebraica, Mexico

1958, pp. 303-308. 7.

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BIOGRAPHY

The author was born in Los Angeles, California, on October 24, 1942.

He attended Harvard University from

September 1960 to June 1964, receiving the degree of Bachelor of Arts, Magna cum Laude.

He began his studies

at M.I.T. in September 1964, was a Research Assistant from September 1964 to June 1966, and was a Teaching Assistant from September 1966 to June 1967. He was employed at the Lockheed Aircraft Corporation during the summers of 1965 and 1966, and did some preparatory work on his thesis while working there.