AN A PRIORI INEQUALITY FOR THE SIGNATURE OPERATOR
by
Antun Qomic
B.S.,
Universidad de Chile, Santiago, Chile
(1973)
SUBMITTED IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE
DEGREE OF
DOCTOR OF PHILOSOPHY
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
June,
(0
1978
1978
Massachusetts Institute of Technology
Signature of
Author.............................
Depar9 menf/ of Mathematics, May 5, 1978
Certified by..
-p.
Thesis Supervisor
-.
-Accepted by.......................................-Chairman, Departmental Committee
on Graduate Students
A
MASSACHUSETTS ;NS3TiUTE
OF TECHNOLOGY
AUG 28 1978
LIBRARIES
2.
An A Priori Inequality for the Signature Operator
by
Antun Domic
Submitted to the Department of Mathematics on May 5, 1978
in partial fulfillment of the requirements for the Deqree
of Doctor of Philosophy.
ABS TRACT
Let X be a C manifold of din 4k, oriented,
X is not C' but locally
with boundary aX, where
of two manifolds with
product
the
as the boundary of
ordinary C' boundary.
(corresponding
For a special Riemannian metric on X
an
prove
we
boundary),
the
near
metric
to a product
is,
(that
operator
signature
the
for
a priori inequality
of
bundle
the
of
sub-bundles
certain
between
d+6 acting
conditions
boundary
non-local
using
forms)
differential
aX. These conditions are defined using eigenon
functions of essentially the tangential part of d+6
X of dimension 4k-l, subjected to
on the pieces of
boundary conditions on the piece of dimension 4k-2.
Using this inequality, we define closed extensions
with finite dimensional kernels and closed
of d+6
images. We study such kernels and give other applications
related to the Laplace operator.
I. M. Singer
Thesis Supervisor:
Professor of Mathematics
Title:
3.
Acknowledgements
With the completion of this thesis, I would like
to acknowledge the help of several persons.
I wish to thank Professor I. M. Singer, my thesis
advisor, for proposing this problem to me, the many
talks on the subject we had, and his personal interest
in my studies.
I am also indebted to many members of the Mathematics Department of M.I.T., particularly Warren Ambrose,
Mark Goresky, Norberto Kerzman, Ed Miller and David Schaeffer.
To Jeff Cheeger of S.U.N.Y. at Stony Brook and especially
Robert Seeley of the University of Massachusetts I extend
many thanks for the time they spent with me.
During my studies I received the financial support
of the Programa LAM of the Universidad Tecnica del Estado,
Santiago, Chile, directed by Jaime Michelow and partly
funded through the Ford Foundation, and the Massachusetts
Institute of Technology.
I am obliged to Kenneth Hoffman,
Chairman of the Department of Mathematics, M.I.T.,
his continued understanding.
for
My thanks also go to
Herbert Clemens, now at the University of Utah, for his
hospitality and concern.
4.
Lastly, I wish to express my deep gratitude to
Warren Ambrose for his encouragement and help from the
time I met him as an undergraduate at the Universidad
de Chile in Santiago.
5.
Table of Contents
Introduction
Part I.
Part II.
Part III.
Operators on faces of the boundary
An a priori estimate
Consequences
Bibliography
6
9
34
50
68
6.
Introduction
In
[3],
Atiyah, Patodi and Singer studied a non-local
boundary value problem for certain kinds of first order
C
elliptic differential operators on manifolds with
boundary, and gave a formula expressing the index of
such problem.
The formula is especially interesting as
it contains two different contributions to the index:
one of the same type as in the index theorem for closed
manifolds
(i.e. the integral over the manifold of certain
characteristic form) and a term related to the spectrum
of an elliptic self-adjoint differential operator acting
on the boundary, essentially the tangential part of the
operator on the interior, that they called the rj invariant
(actually, there is a third term of very
of the boundary
One important operator which
simple interpretation).
fits into the above group is the signature operator, that
is
d+6
acting between two subbundles of the bundle of
differential forms, provided the manifold is taken with
a Riemannian metric that is a product near the boundary.
If we now consider the product of two manifolds with
boundary, say
X
Y, due to the multiplicativity of
and
the signature we can write the signature of
the above result for
X
and
Y.
X
x
Y
using
We obtain an expression
7.
X x Y,
where one term is the usual integral over
and
three more terms that roughly can be interpreted as
contributions from
X x DY,
X
x
Y
3X x 3Y,
and
involving spectrums for differential operators on
these manifolds.
We consider in this work a manifold
boundary
X
X
whose
is not C , but locally it is like the
product case, that is
manifolds with
an index for
Cw
@X = Y
boundary.
U
Y 2 , where
Y.'s are
We attempted to obtain
d+6, acting between the usual bundles
related to the signature, and subjected to certain nonlocal boundary conditions.
As suggested by the product case, we obtain
spectrums and complete sets of eigenfunctions for the
tangential parts of
d+6
on
Y
and
Y 2 , using again
non-local boundary conditions in order to get self-adjoint
This is done in part I, where we also prove
operators.
some estimates that we use afterwards.
When
DX
is
C", the existence of the index and the
formula for it are obtained constructing an explicit
parametrix.
carried out.
In our case, such_ construction cannot be
Instead, we prove an a priori inequality
8.
for the first Sobolev space.
on
DX
The presence of "corners"
forces us to use stronger norm than
H 1/2
on
the boundary to take into account compatibility conditions
at the corner.
Part II consists of the proof of this
inequality, which is the main result of this work.
Using this, we define closed, densely defined
operators
(A+
and
A_),
which are extensions of
on certain differential forms.
d+6
These operators have
finite dimensional kernels and closed images, and we
A_ C A+
get
We didn't succeed in proving that they
have an index (implied by
A_ = A+ ).
If proven, this
last equality would give some indication about the contribution of
methods).
A+
=
A_
X
to this index
(using heat equation
The main problem when one tries to prove
comes from the difficulty in describing com-
pletely the set of restrictions to
in the domains of
A+
and
A_.
X
of the elements
For the Co
case this
can be done, as we remark at the end of Part III,
so
one can prove at least the existence of an index without
using the parametrix.
We also describe some properties
of the elements in ker A+,
as well as other consequences
of the a priori inequality like a weak Hodge theorem and
a set of generalized eigenfunctions for the Laplace operator
on X
(in the sense that they are not necessarily Co on the
closure of X).
9.
I.
Operators on faces of the boundary.
C
oriented and with a Riemannian metric
DY,
boundary
that is
be a 4k-1 dimensional manifold with
Y
Let
a product near the boundary.
We use <
,
> for the inner product of differential
forms induced by the Riemannian metric,
dv for the volume
element and
($P$)
f
=
<$,$>dv
for product in L2 sense.
Y
rinally
*
denotes the Hodge star operator
and what follows, see for instance
(for this
[13], Chapter I7).
Def 1
D
=
-E
(*d
A.
-
(-l)
Pd*)
on p-forms on
Y,
where
p
then it
=p7(-1)
is
immediate that:
A
is a
1 st
order,
elliptic,
formally self-adjoint operator.
Denote by
du
u
the normal coordinate to
is inner co-normal,
Idu 12 = <du,du> = 1.
Y, so that
10.
3Y:
Then we can write near
D = 07- + FL)
where
,
-Pd (*e-du
J=
D=
-
(*dt
on p-forms,
(~Paedu*)
(-l)Rdt*)
-
is exterior multiplication by
e du
du
and
dt
is exterior derivation along directions tangential
to
DY.
6
Can check easily:
acting on
Cw(P(Y)jY)
is also a formally self-adjoint operator.
Prop. 1
aa = -0
Proof:
an immediate calculation shows
D2 = A = Laplace operator on
2 =
Laplace operator on
=
Aa ($+
dU)
tangential to
2
but
2
2
Aa
=
+
Y.
Now:
2
+ ca$
/u
a 2 = -1, so:
(Aa adu,
+ j)a + ;)2
Y
Y
in the sense:
where $,
are forms
11.
-
D/2
32/u2 + a
) + 9a + 1)2
so as the metric is a product near the boundary, we
know
-
A
3 2 /au 2 + Aa, then proposition follows.
Now:
D = a(3/3u -
= a(D/Du +C)
GZ)
C
r
=
-aY
and
(-yb) * = - Z*0* = Z a =
-a Z
SO:
Ca
is a formally self-adjoint lst order elliptic
operator on
C
Co(Q(Y)I3Y).
Then it has a set
{$ }
of
eigenfunctions, orthonormal and complete in L2 sense.
Note:
> 0
if
y.
WCr
i) = -
j
is
an eigenvalue,
OL.
J
= p.
JJ
then:
P
)
(G
so we can assume that for
are of the form
Write
{$
p. /
,a$ }, yi
D = L + M,
where
0, the eigenfunctions
> 0.
12.
L =
-
i)
L* = L,
ii)
iii)
p
*d
on p-forms, and we can check:
L2 = 6d
2
M* = M,
= d6
LM = ML = 0
Then near the boundary, with
L =
GL a/ u + ZL
M =
M 3/3u + cfYlL
aL =
and from the previous equations for
of relations between
L, P
we get a set
aL''''e
Also define on a p-form
V$
L(du):
:
(-1)q*$
if
p = 2q
(- 1 )q*$
if
p = 2q-1
Then by direct computation we obtain
i)
ii)
iii)
Let
v*=
2=1,
DV = vD,
v4
=
p 3 0
VL = Mv
V
(Y
be an eigenvalue of
Z
(they are the
13.
dimensional
same as those of 0) , then have a finite
vector space of
V = V(P)
C
sections:
= {$ IXP = P$}
ML = LM = 0, we obtain
From
= 0.
&,1fL = Pft \,
If
V:
=
Rk
S
=
and similar equations for rT11 .
, ITM: V + V, and also
Then we have that a
on
that
V
As a consequence:
V
, rYM
: V-
V.
as linear operators
can be simultaneously diagonalized, and using
c2 + r2
_
=
2
_
2 I,
and
v t = r1.v, we
arrive to:
there is an orthonormal basis for
{$l'''1?,.,
L4.J
N'
1
*l'
= I'.,
J
.v$. = 0 ,
J
VN}
rf\.$.
V
of the form
so that:
= 0
'Y\.v$. = y(V$
J
J
as for the eigenvalue -p, same is accomplished by
14.
a o }.
{av$.,
But now:
if
$ =
CL($a$)=
=
CL(o+a$)
$,
then
(-as))(p.-a$)
-Pa
+
P$
=-P($+0$)
Then we see:
can find a basis, orthonormal, consisting of
{$|o$
sections for the space of
{$,...,N,
=
Vp,...,V$pN}, and for
Now we look at the eigenvalue 0.
of
Then we see that
AOP$= 0
The solutions
4$= 0,
which
a $ = 0.
2=
4$= 0
will be valid for the
homogeneous degree components of
DY
can take
are the same than those of
C$ = 0
in turn are those of
as
of the form
p$}
-p
C'
$, and recalling that
is a compact manifold without boundary, we see
d$P = 6$P = 0.
if and only if
to the explicit
form of
C
,
Mi, we see:
But if we look
15.
a) $ = 0
if
and only if
Now note:
ck
= a Mr,
-CaL
= rM\
so
= 0
aL'aM
ker2
+ ker
and have:
CL * =~L'
aM
aL'
+
L
Final property that we need is
-M1
aLV = va .
Then usinT
normal forms for anti-symmetric operator3, we get an
orthonormal basis for ker'0
$
forms of the type
Now denote by
B
= E(
-
U
J J
B
j
consisting of harmonic
,
J = 0
J
J
with:
the following operator on
a ) (a.
G..+
J
+
iJ j
.$
j
L2 3Y):
)
where
i)
the
{$
,T$ }
is a basis as previously described,
4.
correspond to
J
for simplicity
ii)
a. = cos 6.,
J
J
a)
p. > 0, and we supress the
.
J
v
j
-
=
sin 6., and besides they satisfy
J
they are the same inside each
V(y ), i.e.
for eiqenfunctions with same eigenvalue.
b)
There is
for all j.
c
so that:
Icos 6.1
> c > 0
16.
2 = cos 26. > 0
a 2
c)
We summarize the properties of the operator
B
in
the following
Prop.
2
= B
and
B*
=
1)
B
2)
Bi = 0
3)
B$
= 0 if
4)
B
is bounded from
implies
B
Bv$
=0
(1-B)a$ = 0
and only if
Hs
+
H
(Sobolev spaces)
and its range is closed.
Proof:
the first three are easy consequences of the form
of our basis, we just do 3) in detail:
B
= 0
($jaj $
iff
iff
( Ip-a 2 a $
iff
(Gyp(-6 $
iff
(1-B)a$ = 0.
+
i$
+
= 0
.ap.)
))
= 0
+ a ca$ ))= 0
for all
j
for all
for all
j
Finally 4) comes from the fact that:
Hs (DY)
=
{Z(a.4.
JJ
Z(Ia.
J
+ b.c y)
J
2 + |b
such that:
J
J
2
J
2s
j
(a2
(cy*
=-a)
1 7
_L /
I1s2j
and
I 112
is equivalent to above sum +
So the
L
proposition is proven.
One can try to characterize the above operators B
from some general properties.
We observe that
B
satisfies the relations:
0
2
2
B
=
BY
a
=
Ba + aB
(i)
(ii)
If we assume some
B,
Hs
continuous from
+ Hs
L2,
and orthogonal projector when restricted to
satis-
fies the above equations, the first of these reduces
the problem to one of linear algebra on each finite
dimensional space
2
{$C
=
2
X2 4}.
J
W(X.)
Both
=
a
V(X.)
and
B
@ V(-X.)
leave
=
invariant.
W
In spite that equation (ii) imposes many conditions
on
B, it is not enough to determine that
B
should have
the previous form.
A more interesting problem arises if we try to
obtain a
level
using these equations just at the symbol
(in those cases when
operator).
to
B
B
is a pseudo-differential
This would give a more invariant explanation
B, we note that the important point is the commutation
relations satisfied by
a, which in turn is a special
18.
case of Clifford multiplication by the conormal du.
is not a pseudo-differential operator,
B
In general
but anyways many important properties of well posed
elliptic boundary value problems (as defined in
Ch. VI) are valid for the pair
[15],
First we note
(D,B).
that we have the inequality:
IV$l
Z+l < C {ILD$jI
for
k >
1
0,
+ IIfPIK + ILBfI!H Z+1/2
1K
denotes norm in
We can see this as follows:
S.
corresponding to
J
=
0
call
for all
H
(Y).
B0
the operator
j.
This is a
pseudo-differential operator of order zero (as observed
in
where it is called P).
[3],
symbol
We can calculate its
(top order) using Seeley's formulas for fractional
powers of pseudo-differential operators, and then we get
that it coincides with the symbol of the operators called
P+
in [15] and
to
D/Du +a..
J(I-B OW
in [10],
Q
Chapter II, corresponding
But then Th. 2.2.1 in
H s+1/2 PY<Cs
{11Dfll s +
[10] tells us:
||$lisl
so in order to establish the inequaltiy we only need
to estimate
B0
by
BBO.
But it is immediate from
19.
the expression for
that there is
B
C > 0
JIBft
S
and the condition b) imposed
so that:
ii JIBB&PW
Ht 0Y)
H
0
Ht
Y)
so the inequality is valid.
Also we can construct a parametrix for
Bp =
boundary conditions
continuous from
a)
BR$ = 0
b)
R
H
(Y)
on
+
0.
H +
D
with
This is an operator
(Y)
R,
so that
DY
is a two sided inverse of
D
modulo
operators which are continuous from
H
+
H+1
(note then they are compact).
The construction of such
complete detail in
R
is carried out in
[3], so we won't repeat it here.
Everything done there is valid in this case
(except the
assertions about the kernels of the operators),
the obvious modifications as we are using
of
B
with
instead
B 0.
Then recalling Green's formula for 1st order
differential operators and Prop. 2,
that is D with domain
3) we see that DB'
20.
Dom D
H
=
(Y)
B$
= 0 on DY}
is self-adjoint, and as in the usual case we get:
D
BJ
has a complete set of C
eigenfunctions {$.},
corresponding to a sequence of real eigenvalues
and the
$ .
J
satisfy
B#. = 0
J
on
{X.},
J
3Y.
Now we proceed to make some calculations that will
be used in proving a priori estimates.
Def. 2
1
HB(Y)
{$
1
H (Y) j B$
= 0 on DY}
with norm:
D $||2
2
||$
+ a jj$112
HB
norms on right hand side are L 2,
and a > 0 is a constant
to be chosen later.
Fix now
f(u)
=
f
in
Cw([0,+))
1
if
0 < u < 1/2
0
if
u > 3/4
>0
otherwise
so that:
Consider the following bilinear functional (the spaces
21.
where it acts will be precise):
9i($,$)
=
f
(f 0)>dv (Y)
<fa$ ,
Y
then:
L
L2
Now:
if
both
$
and
p
belong to
H
B (Y), using the
special form of the boundary condition we see that if we
integrate by parts the expression for
Z
above, the
boundary integral disappears, so in this case:
)=
< D(fa
-
),f$>dv(Y)
Y
and then:
I .($,$)0
1
< (f|
)|1 L2
L2
Lemma 1
there is
E H
s
B (Y)
C = C(f,f') >
we have:
0
so that for all
22.
312 11
I u9 (f$)
2
|JD$|
<i
+
2
C|ti$|
all norms are L
,
2
Proof:
D = a(O/Du +CG), I then if
for
P > 3/4
eH B
and
$ = 0
we have:
IiDf |12 = 111I12 + I IO I12
-Y
As
c Hl1/2 (Y),
Now:
as
the boundary integral is well defined.
= 0, this means on
B
$
= Ea
+ a
Y:
$ J()
then the boundary integral is equal to:
-
2
and as it is
dition
(a.
p
Ea
i
i) , c)
2
. )
preceded
by a minus sign, using our
for the
1
|D$l 12 = 2u
Now set
2
= f$,
,
.
JJ
12 +
then
follows immediately.
S. '
,
12
-=
fi
con-
we see
+
(positive quantity)
+ f'$
and the lemma
23.
In this moment we are able to fix
nition of the norm of
H (Y):
B
we set
a
in the defi-
a > C,
where
is that of the above lemma.
Then:
,
if
e HB (Y)
we have
9(P,'P)I < ILI
2
HB
B
|
"P,) I <
I I|$II
and
L
21
H
HB
0
L
Now we apply an interpolation theorem, all the
notation and norms that appear are as in
[5],
more
precisely we use 10.1 of that reference, with:
A
=
B2
H
(Y) , A
B
A2
B
B
= L2 (Y),
A = B=
so:
A r
H1
2
(Y) =
=
= HB(Y),
then if we set
[L2 (Y)H (Y)]/
[H
(Y),L
2
(Y)]l/
2
, we have
Prop. 3
k
can be extended to a sesqui-linear continuous
functional:
C
24.
P:
H /2 (Y)
I
2
x H
r1 ) I iL
(Y)
I 1/2
and we have
+
lhl H 1/2
HB
HB
We define now a second bilinear functional:
b($,p)
= -
($ID$)
which is well defined if for instance
D
We recall that
with that domain is
,
are in
H
self-adjoint,
then we get:
<
Ib($,i)
L|21'
L
H1
HB
HB
L
jb($,$p)|
B
so the same arguments as given for
to
b,
k
can be applied
and we get:
Prop. 4
b
can be extended to a sesqui-linear continuous
functional:
b
:bBH 1
2
(Y) x H
Ib($,f)l I <
(Y)
H$|
1/2
B
+
II
and we have
Q
II
1/2
HB
(Y).
25.
Now:
interpolation spaces can also be defined as domains
of fractional powers of self-adjoint operators, and for
our case we obtain the same spaces
1/2 (Y) = {Ea.$p.
IX.I
B
where:
[11],
Then we see:
Chapter I).
H
(see for instance
J
{$ }
Ia.I 2
< 0}
J
are eigenfunctions of
DB
and
their
{ x}
eigenvalues.
As consequence of above and density of
H
B
and
, we get:
HB
if
Za
a $,
=
jjj
$p = Zb i
both in
H
B
'
then:
b($,$)
=
-Ex?.b.a.,
J
We need a
J
and then
2
last lemma to obtain the estimate we'll need.
Lemma 2
There is
=
Ea.$.
JJ
C > 0
we have:
so that for all
H /2
'
26.
II
2
II
[L
< E|y
.
2
,H
]1 /
aI
.2
+ cILPI
the norm on the left is as defined in
Proof:
[5].
consider the L2 valued function
= e 6 (1/2-z)
f6 (z)
where
2
L2
2
6
0.
>
f (1/2)
IIf6 (1+it) I
= $
2
1
HB
2
1/2- za $j
Then
and we can calculate
then it
is
If
6
(it) 1122
immediate from its definition
that:
I L$| 2
1
e 6/ 2 (EIxI la 12 + aIMI
<
2
L
[LB 1/2
and taking a sequence
8
+ 0+
2
)
the lemma is proven.
Finally we combine the previous lemmas and
propositions to get our final one, whose proof has
already been done:
we observe that if
a
= 0
for all
x
> 0,
and
L
= Ea i i
then:
in
H
and the
27.
E
E |x.I
ja.1 2X.
X.<O
3
X.<O
j
ja. 1
> 0,
so:
Prop. 5
there is
c > 0
$ =
such that
E
so that for all
H1/2(Y)
B
a.$., we have:
k.<0 3
J
-($IDf)
+
f<fa,
(ff)>dv(Y) + ci
Y
|
22
L
> 0.
As we will see in the next section, we need above
inequality exactly, that is, without multiplying
-(jDp)
by a positive constant, in which case it is a
trivial consequence of estimates for elliptic boundary
value problems.
We present an example to illustrate how the
operator B appears.
see [4],
For notation and the results used,
part 6 and [3],
part 4.
Consider two manifolds with boundary, X and Y, of
dimensions 4k and 4m respectively.
the signature of
X x Y
Then we know that
is equal to (sign X)- (sign Y).
Moreover, as
TXXY
orientations),
=
TX
Y
then
(we use product metric and
28.
Q_ (X
x
Y)
+ Q_
.+
+(Y)
+
+ (X)'Q-(Y)
=
(x)^Q_ (Y)
+ Q_(X)^Q + M
so if we have harmonic forms representing the cohomology
classes involved in the calculation of the signature of
X
and
Y, we can construct using exterior products
X x Y
those necessary for
as in [18],
(we need to use KUnneth Th
Chapter 5 and Prop. 4.9 of [3],
but can be
easily checked).
So consider
4^
where
D c Q+
d4) = 6( = 0
and same for
elements in
(H*(-,a-)
+
pl($)
= -($+Tl)
=
=
Q+ (X)
of
yJ($^A$)
p
where
X
=
with
Q (DX) ,
X
corresponds to
where
(,
for
2
y
is
Y.
the
then:
D
X x Y.
on
X
x
Y
shows:
-
and
$
For the operator
D = D
'
We use subscript 1
H*(-)).
or
E Q+
Y, both representing
on
X
for objects associated to
Then if
$
'
T2 +
(
* )
^ T2(d2+62
explicit calculation
29.
where
if
*
$.
Hodge opt. induced by
=
is an eigenfunction of
D(
Y$
=
D1 $
.$
SJ
S
R)=
on
X.
So:
Di, i.e.
then
= P. ($.^$)*
J
The tangential part of
called
X
D
on
3X
x
aY, that is what we
is:
D ^T2 +
E
*I^T2a2A2
where:
d2 + 62
du
a 2 (du)(3/au + A 2 )
=
is inner conormal to
Now we have:
A2
a2 :
2
and
&+(Y
+-(Y)
A 2 C52 = -C
3Y
near
R+ ()
;Y
T (Y)
+
2 A2
,
A2* = A2
so if:
A 2 4k =kYk,
A 2 $k = Ykk
then
in
Y,
2 = CF2 (du).
30.
denoting by + their components in
As our
$
is in
related to
e
+,
we look for eigenfunctions of
$j^k, where
= $jA"k'
2 .
k 6
e 2 = ~ ll*j1
+(Y).
Set:
2 k
then note:
e1 =
e
e 2 = Yk el
+ yke
2
e
2
-
Using above equations,
and passing to
1
(, = -aD a)
we obtain:
k+y +(Y2 +P) 1/2 )e
1 =-- 'k
T
h=
we get eigenvalues
+p
1 j'k
+(Y2 +2)
1/2)e
j'(kp
where:
C2
-Pk
+
then they satisfy:
2
1/2)2 +
+ (Yk
+ (y2 + P2)1/2
k
I
if
2 +P_ 1/2)e 2
+Y2 +j + 2
1 1/2 e 22}
31.
CY
+P2
.h
=
-(Y
+P
all = h 2
=
1/2h
1h I 112 =
and
and if we express
e
1/2 h
e1
{(Yk+y
+
(y1
(el J
)
+
( 2
in terms of
+ (2 +P
+(Y2 +y12
-
hl,
'2
1/2) h
2
1/2
and:
-
(Y22+)
1/ 2 k (Pj+
ij
VT
kj
+y2)1 /2
k~
Note:
j
jk
2
2 )+y)/
> 0
so:
sign (e, Ie 1 ) = sign
ykI
For simplicity write:
e
= ah
+ bh 2 ,
then note:
a2 + 1)
2
In our case:
from the boundary conditions satisfied by
4) and
ip,
32.
we have that
y
(el1 e)
< 0
and
Yk < 0, so:
< 0
This corresponds to our situation for the proof of
Lemma 1, and as we'll see to condition ii),
c) in the
definition of B.
B
We see that
gonal
space to
Bf =
el,
should be projection on the orthothat
is:
(fjbh1 -ah 2 ) (bh1 -ah 2 )
we already noted
a
our calculation for
to property ii),
+ b
= 1,
(e lae1 )
and
b
-
a2 > 0
as
shows, so this corresponds
c).
Finally we check that
b
stays away from
0,
Note:
b
as
>
b
= -
yk'llj
'Yk1
1/ 2
+ ( 2+
-(2k
Y1jk+1j
)
< 0, so the numerator of
b
(Ik
V/7J
> min {yki
tends to a finite
I
Yk < 0}
> 0
limit > 0,
In more precise way:
and as
is
always
0
IYk
and same for
w.
33.
Cosie
Consider
b
2
21/
2+21yl(
(2y 2+2p
--
=1
=
-2lyl|yl-21pl (
)/2
2c
set
x
jip,
=
c
=
- 2x(x2+y2)1/2
2(x 2+y
2{(x
Numerator
IyI
y =
2
2 +y2)/2
+y2
is:
2{(x 2 +y 2
1/2} + 2{y(x 2 +y 2
2+y2
so passing to polar coordinates we get
b2
=(1 + r sin2
r
1
= T(l
and we note:
were
-r
O cos 0)
sin
cos
sin 0)
+
x
>
|II = min {l
|7I
-r
I-PI
> 0,
i
i.
j
= min {yk<
J
y
< 0}
01
>
|7|I
> 0
1/2
xy}
)/) 2
34.
II.
An a priori estimate
Weconsider now
X
a
Co
manifold of dimension 4k,
X = y1 U
whose boundary can be written as
Y.
J
being in turn
1 ) Y2
so that
C
manifolds with
=
C
1 \Y2 = aY1 = aY2
X
manifold without boundary
Y 2 , the
boundary, and
is a compact
is also compact, and it is
oriented.
On
X
we have a Riemannian metric which is a
product near the
Y.'s
and near
J
Y
to a product of the form
(Y1 n 72)
Such metrics exist,
proven in
as is
n Y,
x
is isometric
[0,3/2)
x
[0,3/2).
(6], for instance.
So we have two coordinates:
x
= normal coordinate to
x2 = normal
coordinate to
Y 2 , defined up to 3/2
Y 1 , defined up to 3/2
so that:
<dx.,dx.> = 6..
1
J
where we denote by
JJ
< , >
the inner product induced by
the Riemannian metric on differential forms.
Now we take
We can write:
d+6
acting on differential forms.
35.
d+6 = oy(dx 2 ) (/ax
d+6 = o(dx)
where the
and
A.
2
( /ax
+ A )
near
Y1 , and
+ A2 )
near
-27
contain the derivatives tanaential to Y.,
a = &d+61
We also have mnaps:
symbol of d+6.
+
2
k+p (p-1)
where
(-1)
T =
*
*
p
on p-forms.
Then can check that the
j = 1,2
pi,
establish isomorphisms of vector bundles, and using
induced metrics and orientations on the
calculation
V
by
x
shows:
<y
$), ()>X
=2
Y.
If:
3
D. =
-
P
3
(*d
d
J
-
(-l)Pd.*.)
33
acting on p-forms on Y.,
induced on
D.
J
=
subscripts referring to operators
3
Y., we can prove by direct calculation:
j
p.-lA...
J
We fix again
]33
f E C"( [Or)
)
as follows:
36.
f(u)
1
if
0 < u <
0
if
3
u > T
> 0
-
2
otherwise
and we fix boundary value problems for each
conditions
B.,
J
on each
stated on the first
D.,
boundary
Y I satisfying the properties
part.
Theorem 1
There is
and
$ =
0
on
C > 0
Y
2,
1 < C{II(d+6) 1k
10
L2
$ e C (X,
for all
such that:
+
we have
+
II401I
L2 LB.
+
1/2 ((2.)
I
where:
denote projection of
eigenfunctions of
that are > 0.
The rest
DB.
J
v'.-l
(1k)
j
on the span of
corresponding to eigenvalues
of this section consists of the proof of
above inequality.
First we note:
can replace all above
1
I|I
by
b
37.
S12,
and
by any
LW
2 (Y
IMI I
for a
Hs M
s < 1.
fixed
Secondly:
if
$
is supported in the interior
the inequality is well known.
of X,
So we see that it is
enough to prove:
I If x 2 )
s
12 < Cs{j
I
(d+6) (f (x1
+ II(d+6)(f(x 2)
+ lI$lI2 + Z
112
)
2
j
2
11 1/2
B
So assume now
and equal to
is
INI
f
0
i
If
on
$
0
is a Co form supported in
on
Y 1n Y2.
(1)
Y1
x
[0,1],
We then have:
equivalent to
2
L2 (Y 1
a2
}
)
Y
) dx
+
f SwI
0
dx 2 +
1$2 1
H (Y XX 2 )
1
I $Il22
L
(X)
we have an elliptic boundary value problem,
with a lst order interior operator
conditions
B1 , as in Part I, then:
A
and boundary
II1 121
1
is
(Y 1x2
equivalent to:
IA 1 1 122
L
(Y1 xx 2 )
+ JIB 1 $|21/2
1
(iY 12)
+ II$ 122
L
(Y
1
xx 2 )
38.
Then we see: in order to prove
(1),
we only need to
estimate
f(x
2
4
122
2
L
+
(X)
f11A
L
0
11
by the right hand side of
isomorphisms,
2x
x
dx 2
(aY 1 n2)
D., etc.
and the corresponding
is clear what we mean and much simpler
to write.
Now we proceed to develop the terms in the right
hand side of (1) .
Recall that
(see for instance
[13],
decomposition of
d+6
l(d+6)lI
12
_
Set now
a* = -a
Chapter IV),
near
;/ax 2
+ 2 Re
(all norms in
(2)
(1).
we should Say
but it
(Y
p)21/2
1|B1(fl
+ f
Strictly
dx
(f)112
0
(3/
and
then using the
Y1:
2 + 11A 1
x 2 A1 )
L 2(X).
p 1 = aA (dx1 ), then
a2 = -
2
39.
L f
<$,A
2 Y 1 xx
1
$>dv(Y)
=
f
<
A
Y1 XX 2 ->21
2
+ f
<$,A
=
>dv (Y)
T2 2
Y
2 Re
f
<
Y
+ f
aY 1
(dv(-)
>dv(Y)
1 2
<pl~ $,
2
A Al,>dv(Y)
x2
>dvO(Y)
x21
denotes Riemannian volume element of the
corresponding manifold).
We integrate above equality with respect to
between zero and one.
Recalling that
$
0
at
x2
x
12
we get
-
f <$,A $>dv(Y)
1(3
Y
=
2
Re
2A
2 A1
1
+ f dx
2 f
0
<P1$,
Y 1 xx 2
The last integral above is equal to:
f <P 1 $,
Y,
>dv(Y 2
2
>dvO(Y)
2
40.
so we get:
II11
11(d+6)$l 12
-
x2
I IA 1 112 -
+
112
f
<O,A $>dv(Y 1 )
Y1
>dv(Y 2 )
f <pl$ ,
Y2
X2
and a similar expression for a form supported on
Y2 x [0,1) .
Applying this to
f (0)
recalling that
1I(d+6) (f(x 1 )
1|
-f Y <'
1,A
4>dv (Y1 )
-
2
-f
I
>dv (Y2
(2
Now we add
s < 1.
C {...}
and
we get:
Y
) P2
a
2
l)
x2)
a(f(x2
2
C {I I$I 121/2
Tx2
+
(3)
>dv (Y2
Il1
lI2 s
(Yj)
We will prove the following:
to the last
V) >dv (Y 1 )
1
HB
an
f (x ) $,
12 + ZlA (f) 112
JX
<4,A
2
Y
= 1,
2 + Il(d+6)(f(x2 ) P)1
1J(f(x
=
f
and
$ = f (x 2 ) $
21
M
}
for
after adding
two lines of the right hand side of
41.
above equality,
the resulting quantity is
positive.
Assuming this claim, we can finish the proof of (1) as
follows:
by the remark preceeding
following inequality implies
B
(fI
1
(Y1
2 <
xx2)
1A1 (f$)112 2
+
(1):
1
21/2
0
W
as the boundary of the manifold
If
and
on
C
L
[0,1)).
E R7
12 2
|121
L
1
xx 2 )
x x2
+1 f
0
L
|L|
DY
x [0,1)
differential
2
dx
(DY xx
I|
E1|22
L
x x2
Then:
2
L
y1
1 x [0,1)
(Y 1 xx
2
)
xx2
x x2'
operator
(Y 1xx 2)
+
(Y
consider
is equivalent to:
+
and
Y
is any 1st order elliptic
H
(X)
}
is any form supported on
DY1 , I|E
1
on
L
Hs(x
X)
In order to prove this, we observe:
x
11C)
22
1
lh$H
+
L
(coordinate
we see that the
d
2
1I
(2),
1x
x x2
42.
I
1I
121 /2(Y
1xx2)
H~
I II H1/2
< C i
< C{above sum}
and after
=
we integrate above inequality with
f(x1f(x 2
.
But now note:
the first
and third term that we
get are already included in
explicit
L.
so we only need to
For this set:
L = L
=
-p A
L
+ 3/3x
,
2
< c(I IAlEI I
1|ILCI22
(X)
then our estimate f or
cIhp12 s
(3),
, here
then is clear that
L2(X)
f IB 1 112
+
I 1-
12
l
)
L 20)
follows after adding
1/2 < s < 1.
Hs M
So we only need now to prove our assertion about
the last two lines in (3).
We use our isomorphisms and see that:
Y,
comes from
$
and we decompose
$
in
if
P near
$+ + $
corresponding to non-negative and negative eigenfunctions
of
D
, we get
irrelevant):
(except for a 1/2 factor that is
43.
{f
Y
<qf+'
+ f <qf _,
Y
-
f <,
Y
(f+)> + f <qf$+'>
Y
1
(f$+)>
1
>
-
f<
1
+ f <qf _,
Y
a
,
(f$
>
Y
we have suppressed mention to
Yl, term for
Y2
in direction
dx1 , this
is of
the same form, and
-1
q = pi Pi~i
is precisely the symbol of
D
can be seen using the equations:
a d+ 6 (dx2
and
d+ 6 (dx )p2
=
and the relation between
A1 .
As
$) was
have that
Now:
E
C
on
X
and
0
on
Y1
Y 2 , we
H 1 / 2 (Y)
using the same arguments of part I and
abstract theorems on interpolation, we obtain the
following inequalities:
i)
If <qf$+'
Y
1
+(f
)>1
< c, 1$+I 121/2
HB
(Y)
D
44.
ii)
If
Y
$
(f$+
axfy1
_
<qf
+
>}
1
<c
$+
1/2
-Y) I 1/2
+ c(E)II$+1
1/2
1/2
HB
B
e > 0.
(here we use
lal |bI < eb2 + c(E) a2)
If <~+,D~+> I
iii)
c<
c
11$ 1/2
B
and according to Prop.
iv)
5, we have for some
- f <$_,DO_> + f <qf _,
(f#_)>
Y
Y
"
then by
(3) , it
c > 0:
+ cI l $p
22
L
>
(Y)
is clear that we have proven:
the inequality:
I | 121
H
(X)
< C{II(d+6) $ 112 2
+
II12 2
L2 X)
+
H.
11 /2
(Y.)
B.
J
J
+ C(C)
L
I
(X)
$ S1/2 2
B.
}
)
j
In order to finish the proof of the theorem, we must
show that the term with
$$
can be eliminated.
Due to
0
45.
the
E,
which can be made as small as desired,
that
it
is
we see
enough to prove:
1I11~ 12
< c{II$I
1/2
IB
2
II
1)
+
M
H
21/2 (
HB
1
}
We suppress subindex 1 and proceed to prove above
inequality.
Set
+ =
E
the
3a
.>
beina the eigen-
I-
functions of
DB.
Then define for
S=f(x
2
)
e
x2
> 0:
2a.x.
JJ3
X. >0
J-
where
f
is as before, then as
is equiH
valent to
1)
2)
3)
C
EIX I
is in
I
2
a I
2
+
||$+
2,
(Y)
we easily get:
M
JY = $ +
1I
|I1
H
H
< (X)
clI$+I IB1/2
Consider now the difference
($-) jY
=
4-E, we get then
(6)
46.
In order to prove (6),
due to the presence of
on the right hand side, it is necessary only
to estimate
z
-
X .<0
E Ix
2
ai
x .ia.12
J J
(_
if
=
Za.(P.
= -
f <(_,D$_>
Y1
=
f<(-,D($-5)>
-
2 + IIf - (E) 112
'
L
-
(7)
d+6:
Using our development of
I (d+6) (f q-y (E) ) I 2
But:
2f <$-(,D((-t)>
Y2
-2f <p
Y2
$-)
2
JT2 (
(8)
-E) >
Now we need the following lemma
Lemma 3
Let
H 1/2 (Y
B1
v E H
Then:
(X)
be such that
v
= vlY
1
is in
47.
V
is in
2vY 2
pendent of
v
1 /2
and there is
2)
2
B1
inde-
such that
2
< c{|I|V||I
1V21 121/2
HB
c
(Y2 )
1 M
+ IIV1121/2
}
HB 1
Proof:
Denote by
on
v1
and
t
a variable equal to
v2 .
in
x2
x
when it appears
Then from results of
[7] , last
section, on compatibility conditions of restrictions of
H 1 (R
elements of
R
x {0}
x
X
R+
X
0
Y
ay xt
1v1 -v2 11
1
Noting that
L 2(Y
X
1
and
B
and have:
< clvI
22
2
1
H (X)
is an orthogonal projection on
|1 l-V21
=1
ay1 xt
lB (V1 -v 2
Chapter I,
|
2 + f
Now combining results from [17],
[11],
{0}
x
), have:
2
ax2 xt
x R+
R+, we get due to local nature o f restrictions
that they apply to our case,
11
R
to
R+)
we see that the norms
DY1 xt
(I-B1 ) (v -v 2 ) 12
section 4 and
ii/2
(Y)
are
48.
equivalent to:
norm 111/2(Y)
+ (f
0
so hypothesis about
f-
I
Y xt
t
0
1
aYxt
v1
implies:
|B v 112 < C,
B 1 (v-v
2<
1
21
HB
but then by the previous lines,
IBiv 2
B 112) 1/2
t ft
2
2)
+
1 1
we see writing
|B 1 v 12
that statement
of lemma follows.
We apply Lemma 3 to
about
C, and recalling observation 3)
C, we get:
III1 2 1 /2
c 144
|$
11
1/2
B 1(Y 2 )
and also to
$-C.
As
f <p1 ($-),
Y2a
($-u)>
< c
-(
H 1 /2 (Y
2
isolating quantity (7)
B
2
from the expression
(8),
II $
and
that all terms are bounded by
121
II
(X)
we see
49.
I $+
21/2
HB
multiplied by constants, that is we have
(Y
proven (6).
But then we have completed the proof of the Theorem.
50.
III.
Consequences
Now we proceed to define an extension of
d+6.
For
this
Def. 3
Set
V = {$ 6 H
I there is
(X, Q+)
{n
I C
C
(X,Q+)
so that
$n
b)
$n =0
c)
$
+
n,+
+
sense
in H
$
a)
on
0
Y
1
n
Y
2
H 1/2(Y
B.
J
in
J
and define:
A+V =
(d+6)i
V c V.
for
We observe:
A+$ = lim (d+6)$i
,
the limit taken in L2 sense, and
the result is independent of the approximating sequence.
Also we see:
a)
C'(X,Q+) c V' obviously, and
51.
b)
If
4 c V, and
H 1 /2(Y)
$
=
V($)
=
and
on
<
Y
a.4 ',
, then
the series
J
s < 1
sense for
H s(Y)
converges in
the eigenfunctions of the
2j
where the
.
are
DB6
The first assertion is consequence of restriction
theorems for Sobolev spaces and for the second see for
instance
or
[2]
[17].
it is well known
We make the following remark:
C
that
(R)
is dense in
as is proved for instance
One can prove that the subset of
Section 2.
in [1],
H (X),
4
=
0
on
Yi1 n2
is also dense in
C"(X)
satisfying
H 1(X),
the proof can be adapted from related results
in
[11],
Chapter I.
So we see that c) is the only
condition that we are imposing in the definition of B.
Prop. 6
A+,
considered as an unbounded operator from
L2 (XQ+) -* L 2(X,Q_),
tor.
is a closed, densely defined opera-
Also: there is
1
1H
<
c{IIA+W
c > 0
2 +
so that for all
|I$
1 2
$ e V:
52.
Proof:
it is clear that
- V.
C
A+
is densely defined, as
We now prove the inequality:
$
tends to
in the definition of
< c{I
V, then we can apply
n 's, so:
our theorem to the
$~nI
en6 C (,Q+)
if
+ I 1n
(d+6)$n
L
l
2+
,+
L 2
11
/2
HB.
J
but by the properties of our sequence, taking the limit
as
n
+
o
we obtain precisely the claimed inequality.
Now we prove that
A+n
{,
n}
V
A+
is closed:
let
$n
'
and limits in L2 sense.
Then using the inequality we have just proven,
n
we see
in H 1 sense, and then
+
in L2 sense, as
operator.
d+6
is a first order differential
To see that
i
can be approximated by a
{$} C C (XQ+) as described in the definition
sequence
of
V, choose sequences
as
m
+
A+n + A+1P=
o
{n,m}}I
C'
so that
$n,pm
with properties required, and choose an
appropriate diagonal subsequence from the "square" of
the
{ nm}.
So Prop. 6 is proven.
n
53.
We now examine the kernel of
define a manifold
Def.
X
so that
X
A+.
For this we
X, explicitly
4
X = X u Y
(-0o,0] U
x
have coordinate
x2
Y2
Y
in
(-coO
x (-o,Q]
and
x1
on
We extend the Riemannian metric and
Y2 x (-m,0].
orientation of
X
to
X
in the obvious way, i.e.
maintaining the product structure we had near
Note that the isomorphisms
Q(Y.
prolong to
J
Q
X {x})
4
Now assume we have
+(X)y .x{x}
Z
=
p..J
a.
in
X.
+
Q(Y.)
over
Q4+X) ly(X)IY
Y.
H 12(Y),
x
where
X.<o
J
Y
is
Y
or
Y 2 , then define on
Y x (-o,0]
-X .x
=
and
$ = p(
accordingly)
(x = x 1 or
x2
3 a.4.
E e
3J3
).
Then is easy to see:
1)
2)
is in C"
(
+ D)E
n Hs(Y x
=
(d+6)$ = 0
0
on
there.
(-o,0])
Y x
(-0o,)
for
s < 1
and then
(-m,0.
54.
lim
3)
H
in
=$
t
t
for
(Y)
4)
.
<
2
x+O
< ce-ax
IIE(-,x)
{IX.I
a = min
so that A. < 0}
J
J
fl e V
Suppose there is
A+n = 0
where
,
L2 (Y
nIY
and
so that:
P().
Define then the following distribution on
a
on
Y
T =
is in
T
x (Or)
1
i
so
on
L
x
Y
f e C (Y x (-o,
=
T[(d+6)f]
(-0,0)
2
We now calculate
so let
Y x
in distribution sense,
(d+a)T
)),
then we have:
<n,(d+6)f>
f
Yx (0, 1/2)
+ f
<4,(d+6)f>
Yx (-0o, 0)
We use Green's formula to reduce integrals to
expressions involving only values over
a)
as
For the first integral:
(d+a)n = 0
on
X,
we see as
f
=
0
Y.
and
fl c H
near
aY
that
it
55.
f k(n,f)
is equal to an expression of the form
where
,
Y
Z
only involves the values of
q
and
f
on
Y
set
Z
= Y x
(i.e. there are no derivatives)
b)
For the second integral:
(-0,-E),
then
<$,(d+6)f> = lim+
f
Yx(-0o,0)
<ip, (d+6)f>
Z
0
Now we use Green's formula for
Z , and by remark 3)
on previous page we can take the limit as
E
+ 0+. getting
the same integral as in a) but with a minus sign due to
the opposite orientations induced on
sides.
Y
from the two
So we get:
(d+6)T = 0
Y x
on
in distribution sense.
(-o,
)
Using this we now state the
following
Prop. 7
The kernel of
A+
is finite dimensional, and its
elements extend to solutions of
p
rn
are inC
H s(X,Q+)
for
(d+f)$ = 0
on
X '
(Y1
Y 2 ).
X, such
s < 1.
As a consequence, the elements of
C
on
ker A+
are
56.
Proof:
the first assertion is an immediate consequence of
the inequality in Prop. 6 and the compactness of the
injection
H
1
(X) + L
2
(X).
The prolongation to
done as we have just described to
such elements are
equation
s < 1
Co
X
is
Y. x (-o,Q], and
J
as solutions of the elliptic
(d+6)$ = 0, and that they belong to
was observed in remark 1) about
3.
Hs
for
So Prop. 7
is proven.
For the image of
A+(V)
in
L (X,Q_), we have
Prop. 8
there is
(
C > 0
so that for all
c V n (ker A+)
in L 2-sense) we have:
H
So:
A+ (V)
(X) < C|A+
L2 X)
is closed in
L2 (X,_).
Proof:
it is a standard argument, if not there is sequence
in
so that
IIA+
nII
- 0,
HlPIK
=
1
,
etc. and this
would contradict the compactness of the injection
57.
H 1(X)
+
(X).
L
We can use Dirichlet principle to solve equation
A
= w,
o c L2 (XQ+)
for
n (ker A+)
.
The arguments
Chapter 7 so we don't give many
are as in [12],
details.
Minimizing the functional on
II(d+6)$ 11 2 -
=
I($)
Denote by
onto
in
Gw
we obtain an element
V
2(wl$)
V r) (ker A+
*
the orthogonal projection of
H
L 2(X,+
ker A+, and recall that all our operators, boundary
conditions, etc. are real.
G
The properties of
can be summarized in
Prop. 9
G : L2 (X,Q)
1)
G
+
L2 (X,Q)
satisfies
is a compact, self-adjoint, non-negative
operator
2)
ker G = ker A+
3)
G(L
2
) c_ V n (ker A+)
as map from
L2 + V,
and
G
is continuous
V with H1 topology
58.
The following eqt is satisfied:
4)
Re((d+6)GwI(d+6)$) =
2
,
$
((d+6)Gwj(d+6)l)
=
for all
o e L
(w-H(w)l)
V.
If all elements
are real:
(w-H(w)1$)
In particular:
AGw =
5)
in distribution sense.
o-H(w)
has a basis of eigenfunctions
G
X
+
complete in L2(XQ+)'
+0,
They also satisfy:
6)
o = H(o) 0
Now consider
A$ . = x.$.
J J
J
can be written as:
o e L
Any
where
{0,1/X.}
and eigenvalues
A+
AGw
(XG+ +
: L
L
unbounded operator, and we look for:
2
*
A+
: L
2
(X,Q_)
L2L(XF2+)
its adjoint in Hilbert space sense.
First we note:
A+GL
2
*
C Dom (A+
(X,Q_)
{0
}
iv
_
0 < X.,
59.
by property 3) of G.
Def. 5
E(X,2_)
$
2 (XQ_)j(d+6)$
F L2(X,Q+
2
(in distribution sense)}
As the formal adjoint of
Dom
On
(A+
we take:
E
(d+6), we see:
is again
(d+6)
c E.
1 2 =
11
$
E
1 2
L
2 + j (d+6)$|
Now we prove an important property of
12
L
2
E:
Prop. 10
The set
W =
{$
s C(X,Q_) I $
=
0
on
Yi t
Y2
is dense in E.
Proof:
note that it is enough to prove that
dense in
Cw(X)
is
E, as this implies the density of H1, and
then use the remark following definition 3.
60.
k : E
So take
continuous linear functional,
-+
any such functional can be written as
,(i)
((I$)
=
,
prove
k E 0
(d+6)
j
+
m
k = 0
(M),
c H1
1
Ho
((c
so
X
M,
c
0
of
and
(
and
we get:
-
=
0
{-n} a C(XQ_)
$ 6 E, then:
as
=
d+6
I(d+6)l)
elliptic, so:
(d+ 6)m llPj
M-X,
on
(ml(d+6)$)
sense.
(X)
(-m I(d+6))
m
and as
(nI(d+6)$f) = lim (n
c H1
Co(R),
on
((d+6)ml$)
Take a general
but
C'(), we
in distribution sense on M.
Now choose
in
on
the extensions by
n,
and
(X,Q_), and
$ sC o().
0
a manifold such that
=
mi
kE
E.
on
Then as
M.
m e H
Assuming
l
denote by
But then
L (X).
M
Choose
to
(TI(d+6)$)
2
with
r
+
=
0
so that
we see:
for all
61.
( jI(d+6) f) =
im
Z($)
then
((d+6)n 1$)
0, as we wanted to prove.
=
Now we look in detail to Green formula for
If
(d+6).
((d+ 6)
Cm(), we then have:
, n
are in
-n) -
((I(d+6)n) = f
ax
-
-
T*
the integral on the right only involves forms of
dimension 4k-l.
and
n
in
E
Now if we assume
in
Cm(XQ+
Cw(X,Q_), we have for homogeneous degree
components:
P =
and
q
*4k-p
-
F q * 4k-q
Collecting terms in above integral using these
equations, we get:
((d+6)
Call
TIn)-
b(t,n)
(Ej(d+6)n)
= 2
*
f
(9)
the right hand side of above equation.
Then we see that by definition of
V
and Prop. 10,
62.
b
extends to a continuous sesquilinear form
b :
V x E+
and clearly
Prop. 11
*
Dom
E(X,_) Ib(-,$)
{
(A+ )
~ 0 on V}
If we now use the following isomorphisms, where
Y
Y1
is
Y22
or
+
_ (Y)
('
$,
-
Q- ()
Y
0 (Y)
take then
so that
$
I-_($)
=+
we neglect those terms
py+-($),
to the integral (9),
+
^
Y.
over
= n
Y
containing the conormal to
f
fl1
2~
Taking now
2
(c+GW)
W+
as they don't contribute
and we get:
1
-
Y
Y
-1
<,v($)>dv(Y)
Y
=-
(1
($
)
(10)
63.
that
is:
((d+6)EIn)
((I(d+6)n)
-
=
-
()
2
J
where
v
(-1 )PE
is
J
definition on
*
J
on p-forms over
p J
Y.
J
(see
on Part I).
B
near
Now recall:
Y
(similar for Y2),
we had the
decomposition:
( /Dx2 + A1)
d+6 = a(dx 2
and
-l
D=
+ A
P
If we do same calculation for
and the iso-
p_, the result is:
morphism
D
Q_(X)
=-l
= -
1_A
1
_
So we observe:
repeated for
Q_(X),
all our results for
can be
Q +X)
but this time we have to consider
as boundary values for the corresponding space
V
the
span of the eigenfunctions corresponding to the eigenvalues that are > 0.
64.
Equation (10) gives the explanation to conditions
for
B
in order that we have 2) of Prop. 2.
From
equation (10) and self-explanatory notation, we have
Prop. 12
)
s V(XI
6
Let
(A+CpI$)
and
$ c V(X,Q_),
then:
=(#A$
so:
A+
is an extension of
A_
We conclude making some remarks on how to prove
existence of the index in the case of
out using a parametrix.
Co
boundary with-
As in any case the parametrix is
needed to obtain the formula giving it, we won't go into
too many details.
The a priori inequalities are valid, we indicate
how one proves
A+ = A_.
We have the corresponding space
dense in it.
and
H /2(X)
E, with
Looking to (11) we see that, as
C"
sections
H 1/2 OX)
are in duality by integration, and the
restriction map from
H 1(X)
tinuous, theelements of
E
+
H1/2 OX)
is onto and con-
define sections over
X
in
65.
H-/2
(X).
form
Then, they can be expressed by series of the
Ea.h ,
J J
$.
the
J
being the solutions of
..
D$. =
J
J J
and satisfying the condition
-l
EX. 1a.1
J
J
2<< 0
The bilinear functional
b($1,$)
b
is then:
=Zb a~kj
jbi
k(j)
where
=ZEb.p.
in
J
and we write
V,
k(j)
to take
if we have any
to
X.
c
<
q.
J
and multiply this
to
1
we see:
J
0, we can find
DX
As
$.
tJ
suffices to extend
on
=
J
on
a.$.
]J
v
E
in
into account.
DX
$ e V
But now:
such that it corresponds
so that
$lax= $..
as constant along the normal to
by a
Coo
cut-off function equal
X.
b(p,$)
such
=
$
0
s V
6
for
have expansions
E
a.5.
J
X >0
it
in
X.
and
$
s Dom A
c.._ E,
66.
Using a variant of Friedrichs' Lemma
conditions),
we can construct a sequence
(d+6)n +
n + V,
so that
and the
Vn
(d+6)p
$
then the a priori inequality says the
sequence in
H1
$
sense, and so
n c H
both in
have boundary values as
belongs to
(with boundary
is
L2
(X)
sense,
above.
But
$n
are a Cauchy
H1
and then it
Dom A_.
As we see, the crucial point in the argument is that
the boundary values of the elements in
set
{cp|$.
corresponds to
In the
C
J
J
X
give the whole
X. < 01.
J
case, the elements of
closed and co-closed.
V
The proof in
ker A+
[3] uses the extension
X, but this can be also proven as follows:
of
D = L + M
D4 = Xc,
so if
DL
same for
=
LM = ML =
0
(see Part I),
we get:
= LD
L2
M,
with
so
M
dimensional spaces
= XL$
and
L
leave invariant the finite
V(X) = {jDe = Xc},
but then by
V(X)
simultaneous diagonalization we can split
VL
are
VM, where
vice-versa for
L
M.
acts as
XI
on
VL,
0
on
in
VM
and
67.
Assume now we have
Z
=
X.<O
a. .
on
DX
c e Q+ (X),
(d+6)$
= 0,
(after using the corresponding
I
J
isomorphisms),
I|dfIl 2
=
then we get using Green's formula:
-(d$6)X
=
(flL$)DX
but now
L4 = LZa
- L $,
so
J
2
1
($L)X
(recall
< 0, but
L*
X7
a
J
= L),
I|d$j
|
JJ
2
|i|
but as the
2 > 0
X. <
J
0, above quantity is
so only possibility is
So
0.
d$ = 65 = 0.
To repeat this argument in our case requires to
L4
examine the boundary conditions satisfied by
B4 = 0
on
DY.
a long calculation as we must use
aL
besides
for the elements
= XA,
D
on
This involves
aD' in
any case in order to obtain a similar division of the
eigenfunctions of
conditions on
B
DB
one would need to impose more
and in that case it does not give
self-adjointness for
DBO
DY
68.
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