_·
____~
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
<
Tf
Tr
1 Y)
satisfies
for any
.
Our construction of a slice is essentially a
generalization of the construction used for the action of
a compact Lie group on a difterentiable manifold (see 4,
p. 108).
Let
M
be a differentiable manifold,
compact Lie group and
A: G X M -> 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<V,V> < 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<I < [s/2] + 1
Similarly, Corollary 23a implies:
(s)
f
is the
binonical coefficient
i
(s)
i
s!
S-)
-20-
Proposition 25
D
(g(f,h))
in the partial derivatives of
such a polynomial will be
a function or
is a matrix of polynomials
f
h .
and
P(g) Daf D h
Each term of
where
P(g)
is
hne coe'ricients in the matrix which repre-
sents the bilinear transformation
g , and where
lal < [s/2] + 1 or I <
f1[s/2] + 1
We now state an important property of
Hs
compares
functions and
Ck
HS(U) , which
0
functions, and which,
when combined with the previous proposition, allows us to
Hs
define
U
of
functions on manifolds.
is an open set of V1 n. Let
Ck
functions from
whose first
k
U
to lk m
Ck(U)
be the set
which are bounded, and
derivatives are bounded.
It is a linear
space under pointwise operations.
Proposition 26:
II
f IIk
Proof:
f
umaxu (
Ck(U)
is complete under the norm
k
Z IDk fl
fl)
usU i=-O0 u
(where DOf
=
f)
Standard, (see 1, pp. 4-5).
We will endow
Ck(U)
with the topology induced by the
It is of course the topology of "uniform Ck
above norm.
convergence".
Lemma 27:
If
(Sobolev)
s > [n/2] + k+l ,
HS(U) C Ck(U)
0
and the inclusion
-21-
is a continuous linear map.
Proof:
See 2,
Remark:
is
H
p.
194 or 20, pp. 465-467.
The above lemma shows that if a function
,
Ck , if
then it is
s > [n/2] + k+l .
because we know that, for all
o
function
sucah
th.h
tS (
in a neighborhood of
is
at
Ck
p , so is
This is so
p s U , there exists some
ot
pf = f
f: U ->Rn
'
p .
n
ITH' C
Un'
IT
Therefore, since
pf
f .
We now discuss the properties of composition of
Hs
functions.
Composition Lemma 28:
Let
g: V ->
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 <
If
[s/2] + 1 ,
Daf n ->
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 I<<
fV
[s/2] + 1 .
As in Case I, we need to show:
Dj n Dvn - D+gn
gn (v)
(v )
f Dg
vn
V1Dagn(v) f
fD v n
D
-
g(v)f
2
fDgj
D
->0
Sra
V tg
n
,_
(v)In v gn
-
a
g(v)
V
f
D ()f
Dg
V g ) vn
-
2
v gn
2
<-Vmas
D
D (v)f Dg
2
v)
g
gD-
D
Dn _
a
JJIV Ign(v) n
v[g ni
2
< max (IDvg - Dgl 2)f
VE
V
Vv
ID (V f l
V
+ max ( IDvgl }f Dg (v)f - D (v)f 2
The first "max" term is bounded, the second goes to zero and
the third is bounded as in case I.
However, for the integral terms we must use the fact
Dgn
that
is non-singular.
gn E C (V)
so
Dgn
is bounded.
(the reciprocal of the
det(Dg
(T
determinant) is also bounded on V , and the bound is uniform
Hence we can assume that
in
n .
We know
f IDaf1 2 < m and
U
JIDa
2
( g(v)fIl2 < max (Idet Dvgl
f ID
Hence
V
1
Idet Dvgj
(U g(v)flD
vvV
>
- 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
<
goes uniand
i
C-L~
1
-25-
-> 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)
_<
K1i e
By proposition 25 and the definition of
fIs
s
Hs
norms it is
enough to show:
There is a constant
K
fU IP(g)I Dae Df 2 < K
such that
II e I1al
l
f lls
where
lal < [s/2] + 1
<
_II
[s/2] + 1 . If lal < [s/2] + 1 ,
or
a el 2 ff Df
max D2
p()2 Imax
ueU Du
U
Dae Dfl
jUIP(g)
f pPDe)
_D~fl
I fl
U ID<I2_I2
C , such that
s
2
and by Sobolev lemma, there is a constant
max IDel < C II e I
.,
u EU
U
Therefore let
find
K
K
in the case
=
(g)l C
By symmetry we can
.
1PI < [s/2] + 1
as well.
The
lemma follows.
Corollary 30:
let
B(e,f)
("."
If e e H (U,
be defined by
MXn)
and
0
ImXn) X H(U,
0
nXp)
HO(U,
B(e,f)(u) = e(u) • f(u)
means matrix multiplication).
B: HS(U
f
Then
np) ->nXp)H(U,
is
a
0
continuous bilinear map.
ýd
r"80
-28-
Remark:
This corollary shows us that the definition of
Hs
a map's being
assume
U, V are neighborhoods of
p e CO (U) , p
1 on
0
be any neighborhood of
such that
outside
U')
so
corollary, also
pp'
p'
E
C(p')
Then p' s CO (U) (extending by
aHS(U) and pp'f . Hs(U) by the
0
0
1 on V N V' . So V N V' and U'
f
is Hs
Using the notation of the above lemma, if
e s Hs(U,E) n c[s/2]+l(u)
linear map
V' C U' and
U'
p'
can be used to show that
Corollary 31:
such that
and
p
1 .
p' r V'-
p
That is,
pf e HS(U) . Now let
0
which is included in U
V
There exists a neighborhood
0
p , is local.
at a point
,
then there is a continuous
pe: HS(U,F) -> HS(U,G)
defined by
te(f) = g o (e,f) oA .
Proof:
required
in
We need only note that the proof of lemma 29
L2
HS(U,E)
derivatives.
bounds on all
s derivatives of functions
and uniform bounds on the first
[s/2] + 1
Therefore the proof works for this corollary
as well.
This concludes our discussion of the elementary
properties of
Hs
functions.
We now go on to investigate
further the effect of composition of maps on the
spaces.
Hs
-29-
a-Lemma1 32:
U,V,g
be as in the composition lemma.
a: H5(TJ) -> H5 (V) be defined by
Let
ag
Let
a (f) = fo g
is a continuous linear mapping, and it is therefore
smooth.
Proof:
That
show that
a_
is linear, is obvious, so we need only
a
is bounded.
find for all
'y such that
that
2
f ID'V~fg)1
<K
For this it is sufficient to
IwI <s
a constant
2
fIDo'fI
Z
=
K'V
IIflii
Ky
such
.B
proposition 2M, we know that to do this we need only find
constants K~ such that
and we can assume
tat < [s/2]
Case I:
where
tat
+t-1
fv
or
IDaf
IS
0
2
D~gl
< [s/2]
IIfII2
< Kar
+1.
< [s/2] + 1
Q is the bound on the continuous linear injection
H (U) C CtaI(U)
0
.Therefore,
le~t
Ka
=
2r
10VjD1g
-30-
e saC
I~
Il
L L2 r- +-
1
0
f IDaf DPgi < max IDg
f V ID(v)f 2
vvV
max (Idet Dgl1 ) f IDuf
< max (JD~g12)
veV
v
Therefore, let
V
v EV
= max (ID1gg12)
K
I f 112
g 2) max [Idet Dvg-
(
max
l
U
vV
S
max [Idet Dg -l)
The lemma follows.
As a dual to the above lemma we have the following:
w-Lemma 33:
g E Ce(V)
Let
V
be an open set in [Rn
with values in
and
~.m ; i.e., for all
k ,
g E ck(v)
HS(U,V) = (f e HS(U)f(U) C V . Then HS(U,V) is
0
0
0
open in HS(U, R n) and w : Ho(U,V) -> HS(U, IRm) defined
Let
by
f -> g o f
Proof:
is a smooth map.
If V is open in
M
n
cO(U,V) = (f E CO(V)If(U) C V}
CO(U)
, it is clear that
is open in CO(U) , because
has the uniform convergence topology.
Hs (U) C CO(u)
But
is a continuous linear injection and
HS(U,V) = Hs(U) n cO(u,V) . Therefore HS(U,V) is open in
0
0
0
Hs(U,
n) . g o f E HS(U, IRm) by Remark 3 following
0
the composition lemma, so the image of wg is in Hs(U, km)
__
-31-
We shall show the smoothness of
wg
in a number of
steps, as follows.
1) mg
continuous.
2) The Gateaux derivative
f e Ha(U,V) ,
dw (f,h) exists for all
h s HS(U, Rn)
0
0
3) dwg(f,h)(p) = Df(p)(h(p))
4) Therefore, Gateaux derivative is continuous as map
from
HS(U,V)
0
to
5) Therefore,
L(HS(Uj
VRn),
0
Hs(U, ikm))
exists and
Dg
DfCg(h)= dwg(fsh)
6) cg is Cý and by induction, smooth.
Proof of steps:
1) Let
-> f
Let
n- f
g fn -> gf
in
S(U,V) ; we must show
0
in
in
Hs(U, 1km) .
As in the composition and
a
lemmas, this means that we must show
fI Dag D fn - Dag D f
But
fuI Dn gD f n -
max IDgl 2 < w
v EV
and f ID fn2)
ag Dpf
%
2
2
-> 0
• max ID glI 2 f
since each derivative of
D f 2 -- >
By definition
0
da
since
fn ->
(f,h) = lim
the limit being in the topology of
t -,O
g
f
ID fn -
P
D f l
is bounded,
in
HO(UV)
I(g(f+th) - g(f)) ,
HS(u, R m)
1
-32-
Therefore it is sufficient to show that
X
(1Da + t h g D
=
lal
sense for
But
and
)
g E C(V, Im)
Therefore, since
(Da+th
- Da)
sense because
is
on
bounded in
U
as above.
- D (u)g)
= D(D~)(h(u))
O < t' < t
is uniformly con-
is uniformly bounded, we know
h
D(D g)(h)
L2
uniformly on
(D(D/g)(h))Dpf
in the
Xt
.
L2
g E Ce(V, Rm)
Daf+th g -> Dag
f
and
uniformly
Dthg
Sf+th g Dh -> Dfg
f2 D h
Therefore
U
U .
on
norm implies
d co(f,h)
sense, so
3)
CO
is
isthg
v .
Now for the second term of
h
s
By the mean value
D(Da(g))
Dfg)D f ->
DAf
norm.
Since
v = f(u) + t'h(u) ,
converges to
(D-a+thg
.
(D(u)th(
,
tinuous as a function of
Hence
CO
v e V , such that
But since
L2
1P1 < s
u E U ,
theorem, for all
for some
converges in the
- Dfg Df)
is bounded in the
h
large,
+t h
(D+thg - Dg)Df + D a
fg
+th gD
=
t
(f
in
L
exists.
I(g(f(u) + th(u) - g(f(u))) = Df(u)+th(u)g(h(u))
by the mean value theorem, where
is uniformly continuous in
so
l(g(f(u) + th(u))
on
U .
By
2),
~)in
on
V
and
0 < t' < t .
h
But
is bounded on
- gf(u)) -> Df(u)g(h(u))
y
U ,
uniformly
.(g(f + th) - g(f)) converges in
v.
Dvg
,
-33-
HS(U, r m)
By the above, it converges to
CO(u, Grk m) or
Dfg(h)
4)
in
Uk m)
HO(U,
HO(U,
j
m)
or
Therefore it converges to
d mg(f,h) = Dfg(h)
Dg: V -> L( Iln .
derivatives.
m)
is
.
is smooth and has bounded
Therefore the map
mD : HO (U,V) -- > Hs(U,L(Rn,,
g
Dfg(h) in
continuous by step 1) .
defined by
,
f ->
Dfg
Let
m )) defined
9: HS(U,L(Rn,
I.m)) _> L(Hs(U, Rn), HS(U, RR
0'
0
by Q(g)(f)(u) = (g(u))(f(u))
It is immediate that
is a continuous linear map
the map induced by
d g
(use Corollary 30).
from
Ho(U,V)
is
9 o aý)g
L(H (U,M n), HS(U, Ilm))
Therefore,
9 o
and
dwg
so by lemma 13,
Dfwg(h) = dwg(f,h)
6)
By the
5
iso9
Also,
to
defines the continuous map
DwI
g
exists and is continuous,
.
steps above,
)Dg: HS(U,V) -> HS(U,L(IR
IRm)) is
0
C1
Also
9
defined above, induces a continuous linear map
9: L(H3(U, Rn),HS(U,L( ,n,
0
,m)))
-> L(HO(U,
Tn), L(Hs(U,
0 n
HS(U, ,m)))
0
defined by
----
Q(M) = 9 oI
Dw2
= 0 o DD
. Hence
w
-·
MM
-34-
C2 .
is
The existence of the
in the same way using
kth
derivative follows
and
D Dk-l
g
(hl
Dkc
Weak
g
"'* hk ) = Dkg(hl
,
c-Lemma 34:
be in
HS(U,V)
such that for all
p
in
U
Let
f e A ,
c-lemma except let
A
D f
f s C[s/2]+1(U)
and
: A -> Hs(U, JRn)
C s/2]+l(u)
q.e.d.
Use notation of the
HS(v) A c[s/2]+1 (V) .
for every
Proof:
"'" hk )
,
and if
f
-> f
be a subset of
is non-singular
.
Then
in Hs(U,V)
and in
cg(fn) -> cg(f) n
cog(A) C Hs(U
,
1T.n)
by the composition lemma.
The convergence property follows from the estimates of
step 1)
fact that
_
of the
Dag
laI < [s/2]+l) and from the
c-lemma (if
is
L2
(if
1
< [s/2]+l).
<I
-35-
III.
Hs
Functions on Manifolds and the Group
Let
M, N
p E M
For
there exist charts
function.
HS(M,N)
which are
Hs
about
p-1 o (f r $(U))
o
If a map is
(U,))
then it is
Hs
Proof:
(U'J1')
p
and
M
to
at
p
f(p)
Hs
N
p
with respect to
(V,*)
(P,-
(V',~')
at
f(p) ,
1
is another pair
o f o $ o (4-o1
o P)*-
'-l1 ($(U) n $'(U')) n f-(4 (V) n ?p'(V'))
p .
*'-1 o *
-1 o
and
on the
')
which is a
4' are
smooth
maps whose derivatives are everywhere non-singular.
using the composition lemma,
and only if
iff
with respect to any other pair.
4'=
neighborhood of
p
M .
Hs
and
at
is an
is the set of maps from
one pair of charts
set
Hs
f is
(V,*)*
at every point of
Proposition 36:
(*')-1 o f o
we say
(U,$) and
respectively such that
If
f: M -> N
be smooth manifolds,
Definition 35:
s
ff o
*-l1
is
',-1 o f o $'
Hs .
is
Hence
Hs
if
q.e.d.
Our next general goal is to define a topology and
differentiable structure on
HS(M,N)
.
To do so we must
use the following:
A chart
(U, ) at
p
is a homeomorphism
4
of an
open set in Euclidean space, U , onto a neighborhood of p
which defines the differentiable structure on the neighborhood (see 13, p. 16).
--
1
--
pro"-
2
-36-
Lemma 37:
Let
f: U -> 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_<1
(IDah i(P)I) < Z C II h
max
pe K
i
i
i
I.I" The proposition
follows.
Now we proceed to define a manifold structure on
HS(M,N) (our method is like that of 6, p. 306).
smooth Riemannian structure on
e: T(N) -> N .
We know that
hood of the zero section of
smaller neighborhoods
in
9'
, such that
restricted to a fibre
system at
tinuous, so
p
e
with exponential map
is defined on a neighbor-
T(N) . Also we can pick some
9 , 9'
with
9
9 the closure of
e [ 9' is a diffeomorphism. e
r
9'
T (N) defines a normal coordinate
(see 15, p. 59).
f(M)
N
Fix a
But
f: M -> N
is a compact subset of
N .
is conHence there
-45-
is an
s
-l (f(M))
in
Now let
e
h
9f
Lemma 44:
*f r Of
Proof:
4'f(h) = e e h .
by
h
e
eh = eh' .
and
e
is smooth.
4
T
f
Tf
is injective.
and
n 7r-(p)
.
c c OTf
B6
is a
a neighborhood of
Qf 2 B6 n Tf
9f
Take
p E M,
Then for all
h, h' E Or
eh(p) = eh'(p)
h(p) = h'(p)
Therefore,
0 .
is injective on the fibres of
Now we show that
fore if
It is clear that
is a neighborhood of zero in
h(p), h'(p) E
Recall
Hs
is
First we show that
since
and define
is injective.
and assume
and
e .
9f = (h E Tflh(M) C 9)
Hs , since
is
contains all vectors
of length less than
Of -> HS(M,N)
1f:
G n r-1 (f(M))
such that
h = h'.
Hence
is a neighborhood of zero in
Tf .
and the injection is continuous.
6-ball about
0
in
for some
Tf .
0
in
B6 n Tf
We shall show that
6 .
For each coordinate neighborhood
Riemannian metric on
CTf ,
There-
N
V
in
N
the
is represented by functions
is a symmetric positive
gk: Vj -> R , where (gkl(p))
definite matrix for all p E Vj . Let *j be the usual
trivialization induced by the coordinate system
(
ai1 •-(p))=
i
1
i
(p,(al,a2 ,
.o. an)))
is
-46-
1/2
be the unique positive definite symmetric
Let gkj
matrix such that (g1/2 ) = gk
and let g-1/2 be its
4-1/2
inverse. Then if X e Tp(N) ,
(gk-/2)
k1/2
) s
L(R
is an element of
-12
(gp
,
X)12
X) = I(x) 2
X, g
IRn)
with inverse
p -> gl/2 is continuous.
and the map
Therefore,
there isa 6 such that if IVI < 6 o Ip
1/2VI <
for all
p e Ki .
j(gl/2X)l <
h
CO Tf
p
Ki ,
.
X e w-I(v)
Assume
(X,X)1/2 < E .
Then
such that
such that
< 6j , then for all
max (lhij(P))
P eKi
(h(p), h(p)) < E .
Let
Therefore if
6 = min [(6j
(the
minimum of a finite set, since there are only a finite
number of
Ki).
Now clearly
of elements
f _D B6 n Tf
h E COTf
B6
such that for any
(h(p), h(p))1/2 < s .
We define
since
consists only
p
in
M ,
The lemma follows.
HS(M,N)
to be a manifold where the set
U
(Gf,of) is a collection of charts for HS(M,N)
fHSB(M,N)
We shall call these standard charts. To be sure that this
definition makes sense, we must check that if
1f
If
1 (4f(gf)
by symmetry of
f
inverse as well.)
and
n *f.(Qf,))
f'
f, f'
is a smooth map.
a HS(M,N)
(Then
we can say that it has smooth
--I PPP"-
Proposition 45:
Proof:
Then
9
f
o *f
E: 9 C T(N) -> 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(
<hij
jh
Wihij
Ui
)
W
and
H s (Wi )
II 113 on Tf defined by (
1 and
so that
II 112
)3 is
s
'
-48-
Proof of the Equivalence:
We know that
1I II 1
II 112
and
C > 0 ,
are equivalent, so it is enough to find a constant
such that II I > II 113 CII l 112 . 11IIi> II 113 because
<hij
h ij I> s > <hij
,
Wi,
hij'
Wi>
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 -> $(h)i
, where
is
o h
be an open set in
Consider thelmap
•
Ci
(Wi'
and
v2 o
we denote
Let
W, •j
hw
i
Tf
v 2 o0j
we denote
such that
C = min [1
and
form defining norms for
h e Tf
Ci
Wi>s < Ci <hij
W1, aihjd
<cihij
h
also by Corollary 31
s
4 is
dk$(h, vl
Q
Ck
to
i
is Ck
Q6, v
h
.
we
1
*
v
k
Then by definition
dk (h, v1 ... vk)
... vk ) defines a contin-
Lk(TfTf,)
S Tf
.
Then by
_:___
-I
-49-
Example 48:
Let
g: T(N) -> T(N)
be a smooth map
which takes fibres into fibres so that
h -> 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
<h(p), h(p)> < 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)
<
5i
, then
peKi
= (hjk)p is positive definite for all p E Ki
Now clearly if s = min (si) and COS2T * has the norm
i
-i(h)p
II 1 II =
h
max
[(1 2 o ?i(#p)l)
then
I h-,ll
everywhere positive definite, or
h e CO
Lemma 59:
< E implies
. q.e.d.
(W s,) is a positive cone in
Ckk
Ck(S 2T*)
(H8 (s2 T*)).
Proof:
If A and
matrices and
a, b
B
are positive definite symmetric
are positive real numbers,
is a positive definite symmetric matrix.
aA + bB
The lemma follows
directly.
Proposition 60:
cone in
For all k,
Ck(S2T*) .
is an open positive
Also for sufficiently large
is an open positive cone in
1
Ck
HS(S 2T*)
s , 9~ s
__~
_
-60-
Proof:
Ck(s 2T*)
The injections
HS(S 2 T*) c CO(S2T*)
Since 97s
)
.
E 9s
Hs(S
2 T*)
Let
Let
B(T(X))
bilinear forms of
metric on
all
X
X
HS(S 2 T*) , and for
= HS(S2T * ) X 6
.
be a manifold based on a Hilbert
be the bundle of bounded symmetric
T(X) ,
(see 19, p. 155).
is a section
y of
p s X , the bilinear form
A Riemannian
B(T(X))
Tp
on
such that for
Tp(X)
definite and defines the given topology of
Let
B(Hs(S 2 T*))
bilinear forms on
T(flS)
Hs(S 2 T*) .
->
-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
<
ai
Then
>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
<
(Vi', i)P
k
,
>k
M
and let
and
p
and the norm
Hs(E)
S E Z fisi
E(Vi k) .
(
Uk
on
ai(P)
Pisi)
has a natural measure
i
on
defined
be the norm
Let
p s M ,
Uk ,
Also on
induced by the coordinate system.
IIk
Vik .
1 .
Each set
for
is a
>i
It is a positive definite inner
>i
because for any
Js(E)
positive for some
II
(pis )
is defined to be zero on the complement of
<
Let
be the
Js(E) , where
smooth positive indefinite inner product on
<
is a
which makes
Vik
an orthonormal set on each fibre.
<
Let
on each fibre.
Js(E)(Vik)
basis of
(S )
E(Vik) . Then the set
a section of
where
(si )
are the basis of sections
Then
2 dxl
lk
a•sU
js k(Da(pf i))
i
..0 dxn
__
-T_
-67-
By corollary 30, there is a constant
(11s 110)2
k
< Ck
k
such that
)2
fU (Da
i
Ck
k
la1)s
2
Let
Bk(S) =
I•l<_s
LI (Daf 1 )
k
k(S < IIS II2k Let s be
of Js(E) induced by S . Fix p = (x
0,... x•)
We shall show that
Bk(S) <
the section
By
Taylor's theorem,
- (Daf)(x-p)a
zI a.
fi(x) -I·I<_•
p
Let
be the monomial
pa
Then
sp
X
a
si a! (D f)Pa
i
= x1 1
since
<U,6> >
-
(si P)
J Ukk ,s> =
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
Mý
-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 <t < 1
f .
fi
If
m > 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
<(k/2
la
ij '
k
In local coordinates
k .
in
HO(S2 T
)
2).
AO(g) = g
a
Note that we have chosen
so the
s
HS(S 2 T * ) c C[s/2]+I(s2T*) , and that
We assume that
(gij) + Pk + Qk "
For any function
h
•
K
-89-
which is a coordinate function of a local section of
SkT*@ S2T
where
k+
,
f
a
Hs
is an
Dk + l (
Therefore
a
If f Da(gj)
h
-1
(and therefore a
•
Dk+1 f Dagi
Proof of
)
2):
But since
a term
is of form
By
1)
S ka(g)n = lalk
Ial =k
f'Da(gij)n
of
Qk(gn ) .
f'
is
C
implies
CO
implies
Therefore,
Pk(gn) + Qk((gn) -> 0
But this and the fact that
that
___A
---
gn ->
0
in
Hk
.
)
ij
Pk(n
-> 0 .
If
, hence
k > [s/2] + 1 , the fact that
k-l-([k/2] + 1)
Qk
k+2
Hk-1 ' Pk (g n
so
is of form
Qk l
in
jal < [k/2] + 1
hence in
f'Dagi
gn -> 0
HS -k C C
and
Similarly if
are
f
nd
D g1ij
SPk
Pk+1 .
a
+ Pk+
Dagi
77 D gij+ f
Dk + l (f ' Dag
+ fh
has form
fDa 1Dagi
of form
q
C[s/2]) function.
z
7 lD gij
la =k Da (gij)) =1al=k+]
Pk
6h
has coordinates of form
We examine
k < [s/2] + 1 ,
(gij)n -> 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 <0
X e Y . Then
Pick
M .
, is a vector field on
a one parameter group of diffeomorphism
E: T(M) -> 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(X n ) -> h(X)
then
as well.
IXnh(t,Xn)
II Xn-X 110
where
But
Cc
II
110
+
- Xh(t,X,p) Idt
11DXn 11011
is
CO
II DXn 110< s < 1 by the selection of
h(t,Xn,P)- h(t,X,p)11
0
norm on
B .
Uii
r
-98-
Therefore,
Sh(X) - h(X n ) O < IIxn
- X 10/(1-11 DX n 110
)
< IIXn- x II0/(1-E)
(*)
From
3(1,x,
k
Dk
3
where
CO
h(X) -> h(Xn)
Therefore
)h= Dk(Id)
th
k
means the
<
DX - DX n
h(Xn)
- DkX
-> 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)
<k
Ck-
)h- Dx(
hl + 1 DXn 1O1D(lk
0o3D(lX,3
(
En
Sn ->
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
>
<
Let
',
D'
corresponding affine connections on
derivatives on
If
Lemma 122:
V'
and
<
>'
q*(y)
be the
respec-
be respectively the
and covariant
T(M)
T*(M) .
X, Y
are vector fields on
1VTTX T
VX'ly= (T)Proof:
7
defined by
and V
and
M,
.Y
is uniquely characterized by the two
properties:
1)
2)
X<Y,Y>'
Y
Y7X
=
2
<VX'Y,Y>'
X -
[X,Y] = 0
(see 11, p. 71).
We shall establish these properties for
-1
Tr1 \ TqX TrY
using the fact that analogous properties hold for 17 .
J
-119-
1)
Xp<Y,Y>'
= Xp<ThY, )qY> (p)
TTI(Xp)(<T•Y,TYY>)
=
= 2 <VTTIX ThY, TY> = 2 <-TI
TrjYY>'
2) (TT)-1VTýX TTY - •-'V•7y TTX - [X,Y]
= Tri-(VTqX TTY -VTrY
TrjX - [TrX, T1Y])
= TU O) = 0
The lemma follows.
r* o D = D' o Tr*
Lemma 123:
*
where
acts on
T*(M) 0 T*(M)
in the usual way.
Proof:
Let
u be a
field which corresponds to
W = 7 - l(W
Da
)
is by definition the map
T
*(y)(
V
Th•jw) = TtUY(W))
the map
(T*)k
the vector
c(Y) = <W,Y> ; 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)
<
and
induced by
Lemma 125:
TE
(T*)k+l
and
7
be the inner products on
>k
r~*(Ty)
respectively.
<a'T>k
a,T E (T*)k
=
* T>k
<7k*
Proof:
Use local coordinates: Let
i1
k ,
dx 1®9..* 9dx
T = Til 0** ik
7 =
yj dx i @ dx j , and
n
xn),...
y n(xl
(xl
S• Xn)
<a,T> k =
1
k aj
"'" Jk 7
is
the matrix inverse of
yi
<TI*(a)2*(-
l 1
***
7
fi
ýxm
where
Yj
iAl
k
11 **
·
k
T) (,Y) I ml
ýx
ik
*•®dx
k
ik k
'
k
T*(7)lm = (7ij
*
i dx 1)
1
i
ax
and
1i
be the map
yl(x I ...
** *
a =
...
.
I kk
ax
~
m1l
ýxmk
k mk
SThelemma follows.()
The lemma follows.
,'
,·
J1 "'Jk
-121-
Corollary 126:
let
<
JS(S 2 T*)
induced on
tively.
Let
Let
g
>
by
and
h
D
y,
2*(g)
, r*(h)
Proof:
be the inner products
j*(y) ,
and
D'
J (ST )
respec-
S2 T* and
induced by
be those induced by
g, h ,
q*(g), I*(h)
hq(p)> = <I*(g)p, Th*hp>
<g(p),
Then
>'
be smooth sections of
g, h be the sections of
and
<
and
This follows from the preceding lemma and corollary.
By the corollary
Dk o D1
Therefore
r*(g) = q*(g)
effect of
r*
l*(g)
where
on each component
JS(S 2 T*) ~
isomorphism
D'n*(g) =
T*
o Dk o Dk- 1 *o
is defined by the
SkT*@ S2T*
and the
S2T* induced by
SkTz
y and
k=0
(Dk]
k=1
follows from the previous lemma.
. Now the corollary
Proposition 127:
p(t)
products
Let
and
be smooth sections of
(
)
l(T*(y))
S2T
(
and
be the inner
HS(S 2 T)
on
, and let
measures on
M
defined by
=..
(g,h) =
f
n),... yn(d x
...
<g,h>p(det 7ij)/2
p
T*()
y and
Using local coordinates, let
xn)
dxl
g, h
(g,h) = (q*(g), q*(h))'
. Then
We use the previous corollary and examine the
Proof:
1(h
)'
a
.
be
T =
. dxn(p)
.Yi
Then
D(g)
-122-
(q*(g), T*(h)) = f<q*(g),
j 1/2
dx 1 *.
q*(h)>p (det(
i ax
p
S.. dxn(p)
J<g, >-1 (p) det(yk) (det yi) 1/2dx
deWs• ! p
1
(det
= f<g,h>
ij
)1/2
p
by the change of variables theorem.
is
(g,h) .
(
S.. dxn(p)
But this last term
q.e.d.
(g,h)
Corollary 128:
Proof :
jp
dx
)
and
=
(
(7*(g),
1*(h))
for
2 T* )
g,h E H(S
)' are continuous and by the
proposition, the equality holds on a dense subset of
Hs(S2T*) .
Corollary 129:
and
(g,h) = (7*(g), T*(h))'
rq E OV s+l
Proof:
Let
7n ->
'q
be the inner product
and
g, h E HS(s2T )
for
p
Also
Since
(l1)*
inI
p(qn*(y))
are continuous,
(Tn) C
on
(I*(g
),
n )* o
is smooth, and
(qIn-I)*
v(g)
Hs(S
D) . Let
2 T*)
T*(h))n _>
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 <
r* is an isometry
Tr E
since
I
and
T y(*)(Tr(Id) (T(Id)
Also,
s+l/I[ ) ,
s+1/Id))) = Tr(Id)
T (d)
s+1
so r*(s) = exp o T(q*)(t) E S
SS ,
Ttr*(t)
Therefore,
s+I/I )) = c(D
i*(4(w
E .
-1
Proof of
2):
i*(s) = 1*(s')
T*(y)
3):
p: v(U) -> U
F
so
A: 0
Therefore
K(u,t) = T(?
II V II <
exp(B) .
(u)*)(t) ,
are continuous.
F
is
Th*(t) = t'
)
7
and let
We shall show
Let
Also if
K: U X S -> 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
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T, aeBoll
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y DBHT1111TTHOC11
saIvoc I4X '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.