IMMERSIONS OF SYMMETRIC SPACES
by
EDMUND FRANCIS KELLY
B. Sc., Queen's University
Belfast, N. Ireland (1965)
SUBMITTED IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF
PHILOSOPHY
at the
MASSACHUSETTS INSTITUTE OF
TECHNOLOGY
June 1970
Signature of Author .
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Departmental Committee
on Graduate Students
ii
Immersions of Symmetric Spaces
by
Edmund Francis Kelly
Submitted to the Department of Mathematics on
June 1i, 1970 in partial fulfillment of the requirements
for the degree of Doctor of Philosophy.
A formula for the second fundamental form of equivariant immersions of compact symmetric spaces G/K in
Euclidean spaces is derived in terms of the Lie algebra
and the classical properties of the immersions are
treated using this formula.
It is shown that if G/K is locally symmetric and
has a 0-tight equivariant immersion in Euclidean space
then it is in fact a tight imbedding and G/K is a symmetric R-space.
Finally an inequality between the ei envalues of
the Laplacian and the Betti numbers of G/K is derived.
Thesis supervisor:
Title:
Sigurdur Helgason
Professor of Mathematics
iii
ACKNOWLEDGEMENT
I am deeply indebted to Professor Sigurdur Helgason,
my advisor, for a mathematical education and his constant
help and encouragement.
I would like to thank my wife for making the "downs"
in my work bearable and the "ups"
enjoyable.
For the first three years, 1965-68,
at M.I.T. I was
supported by a Northern Ireland Ministry of Education Postgraduate Studentship.
A very special word of thanks is due my typist
Gladys Borkowski.
TABLE OF CONTENTS
Page
CD
ii
Abstract
Acknowledgements
Table of Contents
Introduction
iii
iv
v
Chapter 1.
The Second Fundamental Form
Chapter 2.
Tightness
Chapter 3.
51
Bibliography
61
Biography
64
INTRODUCTION.
Interest in equivariant immersions of homogeneous
spaces has increased with revival of interest in the
general theory of immersions.
The most interesting class
of homogeneous spaces is, of course, symmetric spaces.
In this work we classify all locally symmetric homogeneous
spaces with tight (minimal absolute curvature) equivariant
immersions.
We show that the class of immersions exhib-
ited by Kobayashi and Takeuchi [12] are in effect the
only tight equivariant immersions.
In a slightly different vein is the problem of finding
to what symmetric spaces can the work of Frankel [6] be
extended.
One can describe Frankel's method as "Immerse
the space and examine the critical manifolds for nondegenerate height functions."
This work shows that the
extension of Frankel's results to the exceptional groups
for instance will require some modification of method.
An outline of the work follows.
CHAPTER 1.
THE SECOND FUNDAMENTAL FORM.
In this chapter the second fundamental form is
calculated and used to study some classical properties of
the immersion.
Although most of the results are not very
deep the (classical) machinery developed is indespensible
for the rest of the work.
vi
CHAPTER 2.
TIGHTNESS.
We use the term tightness for what most geometers
call minimal absolute curvature to avoid confusion with
the classical meaning of minimal and because most of the
recent developments in the field have been due to topologists who introduced the term.
We develop some interest-
ing properties of immersions which are tight and using
them classify all equivariant tight immersions.
CHAPTER 3.
EIGENVALUES OF THE LAPLACIAN.
Using a formula of Chern-Lashof for total curvature
a weak inequality between eigenvalue of the Laplacian
and Betti numbers is proven.
This formula suggests a
method for searching for tight immersions; this is borne
out by an examination of eigenvalues for R-spaces.
The Lie theory used is based on [7] and as I owe
such an enormous debt to it I do not cite it throughout
but all nonproven Lie results can be found there.
For
geometric definitions [11] was my guide.
I wish to thank Professor Sigurdur Helgason for his
encouragement and many helpful suggestions.
CHAPTER 1.
THE SECOND FUNDAMENTAL FORM.
§0.
Introduction.
Let
M be an n-dimensional manifold immersed in
Riemannian manifold
M
of dimension
N ; for convenience
we shall not differentiate between a point
x E M
and
M as long as there is no danger of confu-
its image in
sion.
a
At any point
x E M
the tangent space
Mx
has
the decomposition
I&= MX e Mx
Mx
where
is the orthogonal complement of
respect to the Riemannian metric on
Y on an open set
composition
for all
if
U c
Y = YT
+
YN
x E U n M .
YT = 0
Let
extend
with
Y
Y
(YT)x E Mx
YN =
be a vector field on
Y .
has the deand
(YN)x E
is called a normal vector field
.
M .
Locally we can
to a tangential vector field, on
denoted by
with
A vector field
0
U NM
where
and tangential if
Y
M .
Mx
F , also
For this reason we shall talk freely
about vector fields on
M
when in fact we mean the exten-
sion.
Let
if
X
and
v
denote the Riemannian connection on
Y
are vector fields on
V Y
=
(
Y)
+ a
M
(XY)
M , then
we write at
x EM
where
and
(VY)x E Mx
X
Theorem 0.1.
ax(X,Y)
(i) The vector field
assigns to each point
x E M
is differentiable, and
on
M
v Y
X
VY
X
which
the tangent vector
(v Y)
X
is the Riemannian connection
given by the Riemannian structure induced from
(ii)
a(X,Y)
The normal vector field
x E M
to each point
X
symmetric in
Hence
E M; 0
a(XY)
the vector
Y
and
Mx x Mx
Proof.
which assigns
a (X,Y) is differentiable,
and is bilinear over
depends only on
>
M .
Xx
and
Yx
C'(M) .
and is a map
Mx
This is standard, cf. [11], page 12.
Theorem 0.1 part (ii) allows us to make the following
definition.
Definition 0.1.
of the immersion
x E M
of the map
Remark.
write
a
Let
for
X
vector field;
The second fundamental form,
M -> M
ax
c ,
is the assignment to each point
given in Th.
0.1, part (ii).
Where there is no danger of confusion we
ax
be a vector field on
then we write
M and
I
a normal
(X))
-(A
= the tangential component of
x
XX
Theorem 0.2.
The vector field
(1)
x E M
assigns to each
the vector
tiable and bilinear over
only on
Xx
gx
and
CO(M)
( ,
)
(A (X))x
, hence
which
is differen-
(A (X))x
depends
Mx
;X
If we denote by
Mx
and by
MXM
A,(X)
and gives a bilinear map
MX
(2)
x
the inner product on
)_
MX
the restriction of
( ,
to
)
Mx
MX
then
( ,
is
)Mx
called the structure induced by
(A (X),Y) M
= (la(X,Y))_
X
Hence
on
A
M and
M
may be regarded as a symmetric linear operator
Mx .
Proof.
Remark.
Cf.
[11], page 14.
When there is no danger of confusion we write
for both
( ,
and
)
(,)M
X
Mx
Definition 0.2.
symmetric operator on
For
Mx
E M~ , Ag
will denote the
given in Th. 0.2.
part (ii).
4,
Definition 0.3.
If
M
is a Riemannian manifold
the immersion is called an isometry if the Riemannian
structure induced by
on
M
M.
If
i
is
we have further
RN
Definition 0.4.
substantial if
of
coincides with the structure
is
is not contained in any hyperplane
f(M)
RN
Definition 0.5.
f: M ->
If
then the height function,
is the function on
M
Definition 0.6.
f: M -> M
is
ca
,
RN
is an immersion
associated with
a E3RN
given by
cpa(x)
and
f: M -> ]RN
The immersion
.
= (f(x),a)
If a group
G
acts on
M
and
Ef
an immersion such that
f(g.x) = g.f(x)
for all
g E G
and
to be G-equivariant.
x E M , then the immersion is said
5.
§1.
The Immersion.
Let
G/K
be an n-dimensional, compact, irreducible
symmetric space with
Lie Algebra of
the standard decomposition.
\ +
We shall always assume that
carries the G-invariant Riemannian structure induced
G/K
by
=
G =
-B
Let
7: G ->
a non-zero
that
r
, where
on
K
B
.
is the killing form on
End(EN)
be a real representation with
fixed vector
e .
(We can and will assume
is an orthogonal representation.)
induces a map, also denoted by
7 , from
Then
G/K
7
into
EN
by
w(gK) = r(g)e .
Lemma 1.1.
The map
gK -> v(g).e
gives an equivariant
immersion.
Although this is well known we sketch a proof since
it is a "sine qua non" of the subject.
Proof.
gK ->
By the G-equivariance of the map
we need only consider what happens at
0
r(g)e
(the origin of
G/K).
Given any
as follows
X E
we get a vector field
X
on
EN
6.
(2f)(x) =
x E EN
for all
(f(v(exp tX).x))jt=0
f E C(EN)
and
If we also denote by
tion of
the corresponding representa-
7
a simple calculation yields
x
= w(X)x
where we make the usual identification of
with its
EN
tangent space at any point.
Now if
we consider
X
as a vector field on
G/K
in
the usual manner, i.e.
(Xg)(p) = d
all
p E G/K
and
(g(exp tX.p))It=0
g E Ce(G/K) , we have
dv(X ) = Xe = w(X)e
dr
where
is the differential of the map
Consider the inner product on
<<X,Y>>
( ,
where
<<,
implies
)
>>
P
r: G/K -> EN
given by
= (w(X)e,w(Y)e)
is the Euclidean inner product on
is
K
EN
invariant so irreducibility of
<< , >> = - cB
where
c > 0 .
But
c > 0
G/K
otherwise the representation would be trivial.
Q.E.D.
As of now we shall assume the vector "e" is chosen
in such a way that the constant "c" in the proof of
Lemma 1.1 is in fact = 1.
Remark.
This leads immediately to the following
lower bound on the dimensions of representations which I
imagine is well known although I have not seen it remarked
in the literature.
Lemma 1.2.
space and
If
G/K
7: G -> EN
is an irreducible symmetric
is a real, class-one representation
then
N > dim(G/K) + rank (G/K)
Proof.
.
An immediate consequence of the following
theorem of Chern and Kuiper [1] and Otsuki [19].
"Let
M
be an n-dimensional compact Riemannian
manifold isometrically immersed in
point
x E M
the tangent space
Mx
2 Rn+p
.
If at every
contains an
n-dimensional subspace with the sectional curvature of any
plane in the subspace < 0 ,
then
p > m ."
Q.E.D.
We now calculate the second fundamental form of the
immersion
lemmas.
r: G/K -> EN .
To do so we shall use two
Lemma 1.3 which gives a local coordinate expres-
sion for the second fundamental form, is well known.
Lemma 1.4 is algebraic and although relatively
simple is very important to the discussion in the next
chapter.
Lemma 1.3.
{x 1 ,...,x
pose
neighborhood
f: M ->
Let
be an immersion.
of
m
M .
in
is the normal component of
Then
identification of ]RN
Cf.
Lemma 1.4.
of
G
)m
(6x (
a((o
)
'
(6 a)m)
under the usual
J
2
Proof.
Sup-
is a local coordinate system on a
n
U
RN
and its tangent space.
[11], pages 17 and 18.
Q.E.D.
Let
7
be an orthogonal representation
with non-zero
K
fixed vector.
(r(X)e, w(Y)w(Z)e) = 0
Proof.
(r(X)e,
Indeed since
for all
Then
X, Y,
Z E
.
First assert that we need only prove
r(Y)r(Y)e)
w[Y,Z]e = 0
= 0
all
X and Y E
.
we have
v(Y)v(Z)e = r(Z)r(Y)e
So we can write
w(Y)w(X)e == 71 [w(X+Y)ir(X+Y)e - w(X)w(X)e - w(y)w(Y)e]
(a)
~ii-- which proves
the assertion.
We know
(r(Y)e,w(X)
= 0
(Y)e)
since the representa-
tion is orthogonal but by (a) this is equivalent to
(v(Y)e, v(Y)r(X)e) = 0
or
(r(Y)v(Y)e, r(X)e) = 0 .
Q.E.D.
Le t
Theorem 1.1.
a
be the second fundamental form
r: G/K -> EN
of the immersion
then at the origin of
a(X,Y) = v(X)r(Y)e
Proof.
Let
U
Let
X1,...,X
n
for all
1
+ ...
+ XnXn
is a coordinate system about
xi 0
)
be an orthonormal basis of
be a normal neighborhood of
Expp (x X
1
X, Y E
G/K
)
0
->
in
0
in
G/K so that
(xl,...,xn)
with
G/K
= Xi
Now
Exp (x 1 X + ...
1
v(Exp (x X
1 1
+
...
+ xnXn) = exp(xlX 1 + ... + XX n )
+ xnXn)
= r(exp(x
1 X1
+ ...
= exp(xlv(X ) + ...
1
To compute
exp(xi r(X
i
2
-x
i6
) + x i(X)
S(0o,0...)
)e
+ XnXn)e
+ Xn (Xn))e
we need only consider
10.
But if
A
and
B
are
rxr
matrices then
1
_2
at
(exp(tlA + t2 B))J (0 ,0 ) =
-t2
(AB + BA)
(This can be proven by expansion in series.)
2"r
6Xi1Xj
= 1 ((X
0(X07(x
)w(x)e+
to
(G/K )
(X )r (Xi)e)
i
which by the proof of Lemma 1.4 equals
But Lemma 1.4 shows
Hence
r(Xi)r(X )e
v(Xi)r(X )e
is perpendicular
so
c(Xi,X j ) = r(X i )r(X )e
For any
X and Y E
3 the bilinearity of
a
gives
a(X,Y) = v(X)r(Y)e
Q.E.D.
Since the classical information about an immersion
is contained in the second fundamental form and since
the form has a simple expression for our immersions one
might expect that the study of the classical properties
would be relatively easy.
To show that this is indeed the case we digress a
little from our main theme and consider the following
two concepts first introduced by Chern-Kuiper [1].
11.
Let
f: Mn -> EN
be an immersion of a compact
manifold in Euclidean space.
Definition 1.1.
if
X
Mx
an asymptotic vector
is
a(X,X) = 0 .
Definition 1.2.
that
in
a(X,Y) = 0
X
If
then
X
and
and
Y
Y
in
Mx
are such
are said to be
conjugate.
To completely describe these concepts for our immersion we have
Theorem 1.2.
For the immersion
r: G/K -> EN
constructed above
(i)
There are no asymtotic vectors at any point.
(ii) If
X EV
are conjugate to
Remark.
then the set of vectors in @
X
which
form a Lie triple system.
By way of illustration of Th. 1.2 part (ii)
it is instructive to consider the sphere
Sn
If we consider
usual way then if
X
as imbedded in IRn+1
is a vector in
tangent space to the sphere
vectors conjugate to
Sn
Sn -l
Sn
p
then the
contains all tangent
X .
In 15] the following is shown.
Consider
Sn = SO(n+l)/SO(n). For each positive integer
an orthonormal basis
in the
fo ''''fm
spherical harmonics of degree
for the space
s
on
Sn
S
choose
Vs
and define
of
12.
Fs (X)
Then
Fs
n-
' "' f m(X))
(fo(X),
X
in
gives an equivariant immersion of
Sn
Sn
in
Em
If we define
k(s) = s(s+nl)
_ (1-k(s))2n
n+2
S
and take
Xi
and
X
orthonormal vectors in the
orthogonal complement of
SO(n)
in
SO(n+l)
then
(w(Xi )r(X )e,w(Xi )w(X j )e) = X/2
= 0
if and only if
s = 1 , i.e.
Fs
is the standard immersion.
Thus part (ii) of Th. 1.2 shows that the immersions
have no conjugate vectors for
Proof of Theorem 1.2.
(w(X)(X)e,e)
(ii)
0
all
X
F
s > 2 .
(i) By Lemma 1.1
in
) .
Perpendicularity is an immediate consequence
of Lemma 1.1 and the fact that
v(X)r(Y)e = 0
implies
dr(X ) = v(X)e
for
(r(X)e, r(Y)e) = 0 .
For the second part we prove the following stronger
result.
Let
X E
define
13.
and
eX = (Y E
v(Y)w(X)e = 0)
A
[Z,X] = 0 ]
=
(Z inA
7X = (W in ~
v(W)v(X)e = 0o
= X qPx
We assert that OX
Z E A be such that
Let
v(Z)v(X)e = 0 .
Thus by Lemma 1.1
v[Z,X]e = O .
dv([Z,X]
Then
) = 0
and
z E Ax
X N
Thus
c
Ax ;
the converse inclusion follows
by reversing the above argument so
W E YX
Choose
Y
in p
X
We can write
=
X "
W= Z + Y , Z E
.
v(W)r(X)e = v(Z)v(X)e + v(Y)v(X)e
= w[Z,X]e + v(Y)w(X)e = 0
But by Lemma 1.4 the terms on the right are mutually
perpendicular.
So
r(W)r(X)e = 0
Hence
X
= AX
The fact that
obvious.
if and only if
Z E
AX
3
EX
X .
is a Lie triple system is now
Q.E.D.
We now turn to
Definition 1.3.
Y
Let
Mn
be immersed in
the
,
14.
Riemannian manifold
normal
Let
a
N .
CX at a point
Then the mean curvature
x E Mn
is given as follows:
be the second fundamental form and
Mx .
an orthonormal basis for
el ...
en
Then
n
E C(ei,ei)
9X =
i=1
Remark.
C
v E Mx
If
then
(v,g ) = TrAv
showing
is independent of choice of basis.
Definition 1.4.
for all
x
in
M
is minimal in
N
if
9g = 0
M .
To show that the type of immersion we are considering is minimal in the sphere Do-Carmo and Wallach [14]
used a result of Takahashi [21] which we state since we
refer to it
again in
"A submanifold
Ch. 3.
Mn
of
SN(r)
(where
r
is the
radius) is minimal if and only if every height function
(see Def. 0.5) is an eigenfunction of the Laplacian with
eigenvalue
-n/r
2
."
However as can be expected from the foregoing, this
can be given a very direct and simple algebraic proof.
Let
7: G -> U(VN)
representation.
as follows:
let
be a unitary class one irreducible
VN
into Euclidean 2N-space
E2 N
<< , >>
be the inner product on
VN
Make
Consider the Euclidean inner product
( ,
) = Re << , >>
This gives us an orthogonal representation of
G
also
15.
r .
denoted by
Pick a
and we have the immersion
Theorem 1.3.
r(gK) -> r(g)e .
The immersions
minimal in the sphere
E2 N
fixed vector "e" in
K
S2N-1 ((n/y) 1
7: G/K ->
2
)
where
E2 N
y
are
is
the eigenvalue of the Casimir operator of the representation
VN
on
of Y
r
The fact that
Proof.
7: G/K -> E 2 N
can also be
regarded as an immersion in a sphere is obvious.
Let
V
affine connection on
tion on
E2 N , v
be the affine connection on
S2 N-1(r)
and
v
the
the affine connec-
G/K
If
X
and
are vector fields on
Y
G/K
then
locally they can be considered as vector fields on
S2 N - 1
E2 N .
and on
X
where
S2N-
a
l (r )
x
c
Y( + (a(XY))
x
x
X
is the second fundamental form of the immersion
E2 N .
->
But
(VY)
X
where
Thus we have
(v ~)
X
+ (E(XY))
is the second fundamental form of the immersion
G/K -> S2 N - 1(r) .
Thus if
immersion
a
G/K ->
is the second fundamental form of the
E2 N
then
1-..-~i _;~
16.
(at(
Y))
(('Y))x
+ (a(MY))
Hence we need only show that the mean curvature normal
E2 N
7: G/K ->
of the immersion
is
perpendicular to the
sphere.
By equivariance we need only consider what happens
0 .
at
g0
Let
0 .
be the mean curvature normal at
n
E r(Xi)(Xi
o =
(Xi )
where
is
i=l
)e
n
= Re E <<r(Xi)r(Xi)e,v>>
i=l
(go,v)
Let
(Ys
Then if
r
.
an orthonormal basis for
all
v E E2N
be an orthonormal basis of
w. r.t.
-B .
is the Casimir operator
n
r
rw(Xi
= -
r(X
i
)
-
w(Ys)w(Ys)
E
i=l
s
n
re = ye = -
rv(X
i ) w(Xi
)e
i=l
in
VN
(oq,v)
where
y
is
real,
= Re <<-ye,v>> = -
which is
perpendicular to
cf.
[23],
Y(e,v)
.
pg. 247.
Hence
S2 N-l(r )
So the immersion is minimal in
trivial calculation now yields
r .
where
So
ye
= -
r = (e,e) .
S2n-(r) .
A
17.
n
-
y(e,e) =
(
v(Xi)Tr(Xi)e,e)
i=l
n
= -
Z
(v(X
i
)e,
= -n
(X i )e)
v
i=l
(n/y)1/ 2
r =(e,e)1/2=
Remark.
.
Q.E.D.
Before closing out the chapter perhaps we
should mention that we have not assumed that the
orthogonal representation
7: G ->
0(N)
is irreducible
in this chapter, as this is unnecessary and in fact much
too restrictive.
immersion of
For instance we have the following
G/K . Let
F
be the space
dim N
of
all eigenfunctions of the Laplacian with eigenvalue
Let
qp,
be the map
(fl(x),... fN(x))
of
F
M ->IRN
given by
}
(fi
where
-X
pc(x) =
is some orthonormal basis
w.r.t. the unique G invariant inner-product.
That
this gives a minimal immersion into a sphere is obvious
from Takahashi's result.
Let
*,(x)
substantial immersion of
G/K
in a sphere such that
be another minimal
coordinate functions form an irreducible subspace of
Then although
cpx
and
*
F
can be regarded as immersions
in the same sphere they are equivalent (i.e. differ by an
isometry of the sphere) if and only if
cf. [4].
FX
is irreducible
However we will see that for the discussion of
absolute curvature there is no loss of generality in
restricting to irreducible representations.
18.
CHAPTER 2.
TIGHTNESS.
Introduction.
§1.
f: M1n -> MN
Let
manifold.
Let
B
the immersion,
be an immersion of a compact
be the unit normal sphere bundle of
i.e.
B = ((x,v)jx E M , v E M
v: B -> SN - 1
We have the map
If
do
SN -
SN -
1
Definition 2.1.
we have ([2]
= 1)
v((x,v)) = v
where
is the volume element of
the volume of
1iv1
,
1
and
CN
is
or [13]).
The total absolute curvature of the
immersion is
T(14,f,R N)
Remark.
If
n
M
=CNCNl-1
v*(da)
B
is orientable Chern-Lashof [2]
showed the formula
T(Mn, fRN)
where
Sm
-
CN
r M( S d e t A1 ddm
CN-1 jM jS Idet Ad)dm
is the unit sphere in
operator on
Remark.
Mm
m
and
A
is the
given in Definition 0.2.
Although the formula in [2] is in a somewhat
I
19.
different form in
formula (21) of [2],
given which is our
in
A
an expression is
terms of moving frames.
See
[24] for instance.
We shall use the following terminology.
S(M)
= (Co functions on
M with no degenerate
critical points)
= # of critical points of index
cp E
n
B(p)
k=O
Bk(p)
min
=
of
(M)
cpE (M)
B(M)
k
( k(CP)
min
(p)
cp E (M)
For any coefficient field
b K (M,K)
K
set
= dimKHK (M,K)
n
b(M,K)
=
Z bK (M,K)
0
Then we have the Morse Inequalities, cf.
5k
( M)
> bk
Definition 2.2.
k-tight if
( M, K )
A function
Sk(P) = Ok(M)
Definition 2.3.
any field
115]
K .
cp E §(M)
is called
e E
is called
•
A function
(M)
20.
tight if
B(cp) = B(M) .
All this is related by means of
Theorem 2.1.
(i)
and if
If
p
(M)
cp E
Let
is k-tight all
then
k
then
p
is tight
the Morse inequalities are equalities for some
coefficient field
K
and some function
r E §(M)
then
this conclusion can be reversed.
(ii)
If
Proof.
p
(i)
Is obvious
(ii)
See [16].
Theorem 2.2.
Mn
is tight then it is 0-tight and n-tight.
Let
in Euclidean space.
inf T(M,f,
I
Q.E.D.
be the set of all immersions of
Then
RN)
= B(M) > E B(M)
fel
Proof.
K
See [13].
Definition 2.4.
total curvature if
Q.E.D.
An immersion in
RN
has minimal
T(M,f, IRN) = B(M); such an immersion
will be called tight.
Remark.
It is well known that not all manifolds
have tight immersions,
e.g. the exotic sphere does not
have one [13].
We shall need the following well known lemma.
21.
Lemma 2.1.
is tight if
pa
(Kuiper [13])
An immersion
Mn
RN
in
is tight for all height functions
pa
with non-degenerate critical points.
This leads to
Definition 2.5.
pa(x) = (f(x),a)
An immersion
f
is k-tight if
is k-tight whenever it is non-
degenerate.
Theorem 2.1 shows that an immersion which is tight
is also O-tight, but Banchoff, cf.
[14],
has shown that
the reverse is not necessarily the case; however we will
see that for equivariant immersions of symmetric spaces
they are equivalent.
A theorem of Chern-Lashof [3] shows that if
f: M -> IN
is
an immersion and
iof: M ->IRN+ 1
immersion induced by the inclusion
T(f,M, ]N) =
T(i f,M, JRN+l)
.
i: ]RN ->
EN+l
is the
then
Also total curvature is
invariant under affine transformation [13]
so the search
for tight immersions can be restricted to the study of
substantial immersions.
22.
§2.
Reduction of the Problem for Homogeneous Spaces.
We now prove two theorems the second of which proves
a conjecture of Wilson [22].
In fact [22] contains a
a very different
particular case of the theorem proved in
manner.
Theorem 2.3.
space and
of
7
G ; if
sion
r
is
p(p)
is a compact homogeneous
reducible
7
= pt1
p
p
and gives an immer-
which is O-tight and substantial
p: G/H ->
gives an immersion
the immersion
Remark.
G/H
is a class-one orthogonal representation
7: G/H -> Ev
then if
p
If
E
, then
is 0-tight.
need not give an immersion; for instance
could be the trivial representation of
G .
Before proceeding with the proof of Th. 2.3 we recall the two-piece property, cf.
Definition 2.6.
Then
f
(m E M If(m)
Lemma 2.1.
f
f: M -> EN
be an immersion.
has the two piece property if given any hyperplane
H c EN .
The
Let
[14].
[14].
'
H)
Let
has at most two components.
f: M -> EN
be an immersion.
is O-tight if and only if it has the two-piece
property.
Proof of Theorem 2.3.
Let
E
p
and
E
P
be the
23.
representation spaces for
Since
7: G/H -> E.
vector
p
and
ut
respectively.
is substantial then the
e = e
p
+ e
0
e
O
e
p
H-fixed in
E
H-fixed in
E
gives an immersion of
into
G/H
Suppose
p
show
satisfies the two-piece property.
any
fixed
can be written
e
p
H
p
E
.
We
Since
r
satisfies the two piece property given
v E E
,
(p E G/H
#
((p),v)
has at most two
c)
components, for any constant c.
Write
v = v
p = g.0 .
Then
In particular if
vp E Ep ,
+ v
.
v
EE
Ep
and
v
(v(g)e,v) = (p(g)ep,v)
we consider
v
= 0
(gH E G/H I (p(g)e ,Vp)
E E
+ (U(g)e
, and
.
v)
then for any
,
c)
has at most two
components which is the two piece property for the immersion
p: G/H -> E
.
Lemma 2.1.
Q.E.D.
Hence the immersion
Corollary.
space and
G
r
Suppose
G/K
p
is 0-tight by
is an irreducible symmetric
an orthogonal class-one representation of
such that the immersion
v: G/K -> EN
is 0-tight.
Then there is an irreducible orthogonal class one
representation
w'
of
G
such that
w': G/K ->
EN
is
0-tight.
Proof.
There is no loss of generality in assuming
wft
24.
r: G/K -> E N
i.e. a
v
substantial.
with
For if not then there is
(v,w(g)e) = 0 .
vector so there is a
G
If
e
is K-fixed
invariant space
with
Ev
(E,G/K) = 0 .
Suppose
EN = E PE
P
.
u
Then since immersion is
substantial, we can write as in Th.
Then by proof of Lemma 1.1
or
I
I
0
w(X)ep
either
X E
all
p .
p or
r(X)ep = 0
Assume
p
E
sion
where
G/H -> E.,
all
+ e
.
X E )
) ; so we get immersion from
peat the process for
':
e = e
2.3
gives immersion.
We re-
and eventually we get immerr'
is
irreducible.
repeated applications of Th. 2.3 show
r'
Then
is 0-tight.
Q.E.D.
Remark.
Henceforth we assume all representations
are irreducible unless explicitly stated otherwise.
Theorem 2.4.
space and
G
r
Let
G/H
be a compact homogeneous
a class-one orthogonal representation of
(not necessarily irreducible).
7: G/'H -> EN
by
w(gH) = r(g)e
If the map
where
e
is
the
H-fixed vector, gives a 0-tight immersion, then it is
in fact an imbedding.
Proof.
if
Let
Y=
+J9
for any
X E 7
Suppose
g 0 H
with
e
be the H-fixed vector in
RN .
is the direct sum decomposition
r
since
7
Then
r(X)e
is an immersion.
is not an imbedding.
w(g)e = e , i.e.
if
He
Then there is
is the subgroup
#
0
25.
leaving
e
fixed then
H
H
properly.
But since
He
is also
H /H
G/H
is
is
is
7
; thus since both groups are compact
finite.
If
index
[H,H ]
an m-fold covering of
G/H
is
.
m
say, then
Denote this cover-
.
ing by
We can regard
where
an immersion the Lie algebra of
'(gH
7: G/H -> R
e
) = v(g)e .
by
as imbedded in IRN
G/H e
by
So we can factor the map
.
7 = Vo
Now consider the height function
cp(x) = (r(x),a)
pa
on
G/H .
x E G/H
= ('or(x),a)
(X)
where
pa
d' o dT .
is the height fn on G/H . Thus dp =
So the singularities of Cpa occur "above"
the singularities of
a , cf. Fig.
*
FIGURE (i)
(i).
_I
_._;__ i___ _~
26.
Since a singularity is a local phenomenon the
singularities have the same type and index on both
G/H e
and
But
Ne
pa
G/H .
G/H e
there is
Thus for any non-degenerate
is compact.
at least one critical point of index-0
(namely the minimum).
So
pa
has m-points of index-0.
Contradicting 0-tightness.
Hence
H = He
and
r
is an imbedding.
Q.E.D.
27.
3.
The Second Fundamental Form of 0-Tight
Immersions.
The following theorem is a useful improvement of
Theorem 4 in [13] and represents more of a change in
point of view than anything else.
Theorem 2.5.
If
f: M -
RN
is a substantial
0-tight immersion then there is an open subset
such that
M
a: lmMm
Proof.
pa
a ERN
Let
is non-degenerate.
mum at
x0 E Mn .
we can assume
-> M
is
U
onto for all
of
m EU .
be such that the height function
Assume
CPa(x)
attains its maxi-
Since tightness is translation invariant
f(x ) = 0 .
Then the function
CP-a(x) = - (a,f(x))
has a non-degenerate critical point of index
If
0
#
ax
0
xo
is not onto then by Lemma 1.3 we can choose
z E 40such that
Fz(x) = -
(a-z,f(x))
has a non-degenerate critical point of index- 0
Assert
+ve
at
and
aek
-ve
Fx(x) = - (a-Xz,f(x))
at
assumes both
values.
The function
h(x) = (zf(x))
(a,f(x))
x
is not constant
.
28.
I
f
since
is a substantial immersion.
such that
F
= -
h
takes values >
(a-Xz, f(x))
Let
Thus there is
and values <
assumes
+ve
and
X
Thus
-ve
values.
w = xz - a .
Then
cp (x) = (w,f(x))
has a non-degenerate
critical point of index-0 at
Assert we can choose
x
w'
.
in IRN
such that
has non-degenerate critical points and
critical point of index- O0 near
x0
cpw
w' (x)
has a
which is not a true
minimum.
Since
pw(x)
has a non-degenerate critical point
of index- O0
at
(u 1 ,...,un)
on an open neighborhood
there is
xo
a local coordinate system,
U
of
x o = (0,...,0)
such that
2
CP (x) = u
2
+ ...
+ un
on
U .
Consider the "sphere"
S(r) cU given by u + ...
+ u2
1
Un
2
2
r
Then p (x) = r
on S(r) . We can choose w'
in any neighborhood of
w
such that
pw'
has only non-
degenerate critical points
lipw -
IW,I (f(x), w- w') IM
1M=
< Alw- w!
where
A = max !if(x)1!
xEM
2
Choose
w'
wi th
< r
Iw-w' 11!p~
n
Then
The
2
Icpw-w i < - - •
7
29.
f(r) .
has a minimum in the closed ball
cpw,
that this minimum does not occur on the sphere
We assert
S(r) .
But this is easy since
, (x o ) = 0
2
and
So the minimum on
index- O
T(r)
takes
pw
that we can choose
values.
minimum of
So
for
x E S(r)
is in fact a critical point of
w' .
of
Since
-ve
>-
(x)
cp
and
+ve
w'
-ve
such that
values it
@w'
takes
is
clear
+ve
and
Hence the point will not be the absolute
cpw
cp,
"
is a non-degenerate height function with
two critical points of index- 0, contradicting 0-tightness.
Hence
a
must be onto at
The fact that
U
xo
of
of
a .
is
a
.
is onto in an open neighborhood
a trivial consequence of the differentiability
Q.E.D.
Corollary 1.
Cf. Th. 4 in [13].
N-
Proof.
Mxm
x
to
n < 1 n(n+l)
--2
Trivial since
Mx.
Corollary 2.
a
.
is a symmetric map from
Q.E.D.
Let
M = G/H
be a homogeneous space
30.
and
w
a real class one representation of
such that the imbedding
substantial.
Proof.
The
a
Obvious.
r: G/H -> EN
G
on
EN
is 0-tight and
is onto everywhere.
31.
§4.
Symmetric R-Spaces.
The theory of R-spaces, as we need it, is scattered
throughout the literature so in this section what we
need is organized with outlines of the main proofs.
We
do not define a general R-space but give a somewhat
ad-hoc definition of symmetric R-spaces since that is
all we need.
Let
Z E4
? be a real semi-simple Lie algebra and
such that ad Z is semi-simple with real eigen0, +1 .
values
Theorem 2.6.
=,=
+ 5
such that
Proof.
tion of
Then
4
Z E
.
Cf. [10], Ths. 2 and 3.
z =
Let
There is a Cartan decomposition
-l
+
+
;?o +
and define
be the eigenspace decomposia: 4 -> Z
a(X+Y+W) = -X+Y-W ,
X E £oi
a
automorphism.
Let
is
an involutive
a'
,
Y E ;eo
Z E
.
' + 3' be any Cartan decomposition with
4 =
involution
.
Then if
-B(X
Y)
B a' (X,Y)) =
= -B(X,a'Y)
B
a linear map by
X
nd Y
X and Y E e
C'
is a symmetric, positive definite, bilinear form on
32.
;
aa t
and
is self-adjoint w.r.t.
B
at
.
Now it is
almost standard that there is a Cartan decomposition
+ 6
S= e
a
with involution
such that
a
and
a
commute.
Z E 6
that
direct sum.
(L,G)
L
ad Z
To see
is sym-
Q.E.D.
Ba .
Now let
be a pair associated with
(;,a)
has no center.
Theorem 2.7.
(i)
N0
it suffices to show that
metric w.r.t.
such that
z
o N
Zo =
Thus
G/K
K = fg E Glad.g(Z) = Z3 .
Let
Then
is symmetric
p: G/K -> B by
(ii) The immersion
cp(gK) = adgZ
is
Proof.
Int(Z)
tight, equivariant.
(i)
and
Zc
has no center,
Let
Lc
Ad : L ->
be the complex Lie group
is the complexification of
exp(irZ)
Let
by
L
is an analytic isomorphism onto so we shall assume
L = Int(Z) .
where
Since
E Lc
where
Z .
Then
Then
First we show
Ad c(exp iwZ) .
;e
e2
is
L c Lc
i = 4--1 .
e denote the inner automorphism of
exp(irZ) .
Int(,c)
= Id
.
We assert
stable in
Ac
G
under
Lc
defined
is
e-stable.
33.
Let
and
W E-
WI E J1
, W = Wl + W + W1 , W_1 E -
' W E o
in the notation of Th. 2.6.
Ad(exp iwZ)W = Ad(exp irZ)W_ 1 + Ad(exp iVZ)W°
+ Ad(exp iwz)W
1
= e
W_l+
W o + ei W
= Wo - W1 - W1 E ;.
Now
a Ad(exp irZ) = Ad(exp iraZ)o a
= Ad(exp-
iwZ)o a
= Ad(exp irZ)
So
is
stable under
IG
We let
Let
(K
K8
)o cKc
be also denoted by
Thus
K
where
K
.
(KS)o
So
G is 8-stable.
8 .
be the fixed point set of
the identity of
(ii)
Ad(exp irZ) .
o a
8 .
Then
is connected component of
G/K
The equivariance of
is Riemannian symmetric.
p
is obvious.
Tightness is proven in [12] Th. 3.1 for an even more
general type of space.
Definition.
Q.E.D.
A symmetric homogeneous space
G/K
is
a symmetric R-space if it can be constructed as in Th. 2.6
and Th. 2.7.
34.
§5.
The Fundamental Lemma.
We now examine the implications of Th. 2.5 for
equivariant immersions of symmetric spaces.
Let
G/K
be an irreducible symmetric space,
an irreducible class-one representation of
e .
with a K-fixed vector
Denote by
w(gK) = r(g)e , of
G/K
fundamental form is
onto we have
in
EN .
7
G
on
7
EN
the immersion
Then if
the second
EN= T + T
0
0
T o = fw(X)e ; X E
where
To = linear hull of
fr(X)r(X)e: X E
We shall need the following lemma.
Lemma 2.2.
of
T
If
is a real orthogonal representation
r
with a vector
e
0
annihilated by
(r(X)r(X)e , w(Z)r(Y)r(Y)e)
for all
X, Y
Proof.
Z
We know
We can rewrite
So
and
in
then
= 0
.
(r(X)r(Y)e, w(Z)r(X)r(Y)e) = 0
r(Z)w(X)r(Y)e = r[Z,X]r(Y)e + r(X)r(Z)r(Y)e.
(w[Z,X]w(Y)e,w(X)r(Y)e)
But the first term is
+ (r(X)7(Z)r(Y)e,r(X)r(Y)e) = 0.
zero by Lemma 1.
So
35.
o =
(r(X)r(Z)r(Y)e, r(X)r(Y)e)
= -
(r(X)r(X)r(Y)e,
r(Z)v(Y)e)
=-
(r(X)r(Y)w(X)e,
r(Z)v(Y)e)
since
w(Y)r(X)e = r(X)r(Y)e
w(Y)r(Z)w(Y)e)
= (w(X)r(X)e,
as above
= (r(X)r(X)e, w(Z)r(Y)r(Y)e)
.
Q.E.D.
Now let
Z = E
EN
Give
;
the following algebraic structure.
(i)
X,Y
(ii)
X
in
(iii) u,v
7
u
in
EN
-B([u,v],X)
Lemma 2.3.
If
in
[X,u] = -[u,X] = v(X)u
EN
then
.
as in
IX,Y]
:
in
[u,v]
is in
= - (v,w(X)u)
G/K
for all
EN
X
G
u E EN
just note
.
r
a
giving imbedding
Z =
Anti-commutativity.
+ EN
For
into a Lie algebra.
X
and
Y in
then anti-commutativity is inherited from et.
and
in
then if the second fundamental form is onto
the above operations make
Proof.
where
is a symmetric space and
class-one orthogonal representation of
T: G/K ->
V
then it is defined.
(u,w(X)v) = -
For
u
and
For
v
in
(v,w(X)u) .
So now we need only check the Jacobi identity.
Unfortunately this must be done case by case.
X E
EN
V
36.
Case (i).
X, Y
and
Z E
then it is inherited
2.
from
Case (ii).
and
X, Y E
EN
u
then
[X,[Y,ull]] + [u,[X,Y11 + [Y,[u,x]]
= r(X)r(Y)u - 4[X,Yju - 7(Y)w(X)u
=0
Case (iii).
XE V,
u and v E EN
Let
Y
be in
Then
(Cx,[u,v]],Y)
+ ([v,[X,u]],Y)
([u,v],[X,Y]
+ ([u,[v,x]],Y)
([v,r(X)u],Y)
- (lu,w(x)v],Y)
(v,r[X,Y]u) - (v(X)u,r(Y)v) + (7(X)v,r(Y)u)
=
S(v,7[X,Y]u)
=
0
+ (r(Y)r(X)u,v) -
(V(X)r (Y)u,v)
.
Now before we consider
u, v and w
in
EN
we
develop a few preliminary results.
If
for
Z E
X
and
Y
are in
P
([r(X)e,r(Y)e],Z)
.
([1(X)e,r(Y)e],Z) = -
Thus
consider
([v(X)e,r(Y)e],)) =
only consider
Z E A.
(v(Y)e, v(Z)r(X)e)
by Lemma 1.4.
So we need
.
37.
= (-r(Y)e,r[Z,X]e)
-(7r(Y)e,7(Z)7(X)e)
Thus
[(X)e,r(Y)e]
=-
(Y,[Z,X])
=-
([x,Y],Z)
= -
[x,Y]
(a)
Now we can consider
Case (iv).
u, v, w
in
T
, u = r(X)e , v = r(Y)e ,
w = r(Z)e .
+ Iv(Z)e,[r(X)e,r(y)e]]
[v(X)e, [(Y)e,r(Z)e]]
+ [v(Y)e,[v(Z)e,r(X)e]]
= r[Y,Z]v(X)e + w[X,Y]v(Z)e + r[Z,X]r(Y)e
- r[X,[Y,Z]]e-r[Z,[X,Y]]e
= 0
- w[Y,[Z,X]]e
by Jacobi identity on
Case (v).
u, v in
To0 and
w
in
By Lemma 14we need only consider
r(Z)r(Z)e
by (a)
with
Z E
TO
w
of the form
so
[u,[v,w]] + [w,[u,v]] + [v,[w,u]]
= [r(X)e,[((Y)e,r(Z)r(Z)e]]
+ I[(Z)r(Z)e,[(X)e,w(Y)e]]
+ [w(Y)e,[
(Z)r(z)e,r(X)e
u = w(X)e ,
where
=-
]]
v = r(Y)e
r([r(Y)e,w(Z)r(Z)e] )r(X)e
+ r[X,Y]r(Z)r(Z)e
+ T([r(X)e,7(Z)r(Z)e])r(Y)e
(b)
38.
Assert
[r(X)e,r(Z)r(Z)e]
For any VE
is in 6
all
X and Z in
9
I
-
([ (X)e, (Z)v(Z)el ,w)
(v(Z)r(Z)e,r(W)r(X)e)
=0
by Lemma 1.4 since
if
WE
Y(W)r(X)e = r[W,X]e
E T
So we can
write
v([r(X)e,r(Z)r(Z)e])r(Y)e
Now let
Xi )
= r(Y)r([v(X)e,7(Z)v(Z)e)e
be an orthonormal basis for
[I(X)e,w(Z)r(Z)e]
.
Then
n
E ([i(X)e,7(Z)r(Z)e],Xi)X i
i=1
Thus
v([v(X)e,w(z)r(Z)e] )e
n
=
([r(X)e,7(Z)r(Z)e],Xi)w(Xi)e
i=l
n
(r(Z)Y(Z)e,(X
i=1
i
Z (v(X)r(Z)r(Z)e,
-
v(Xi)e)v(Xi)e
(X)r(Z))(z)e
Since by Lemma 2.2
a basis.
)r(X)e)r(Xi )e
w(X)R(Z)r(Z)e E To
Substituting in (b) we get
and
v(Xi)e
is
39.
r
[u,[v,w]] + [w,[u,v]] + [v,[w,u]]
= -
r(X)r(Y)r(Z)w(Z)e + r[X,Y]r(Z)r(Z)e
+
(y)r(X)r(Z)r(Z)e
0 .
Case (vi).
u
in
T o , v and w
u = r(X)e , v = w(Y)r(Y)e
in
Let
T
Then let
, w = V(Z)r(Z)e
J = [u,[v,w]] + [w,[u,v]] + [v,[w,u]] .
Assert
J E T
J = [r(X)e,[v(Y)r(Y)e,w(Z)r(Z)e]]
+ [v(Z)v(Z)e,
[v(X)e,w(Y)v(Y)e]]
+ [v(Y)r(Y)e,
[r(Z)r (Z)e, r(X)e]]
We saw in the course of the proof of case
[r(X)e, r(Y)r(Y)e]
W be in
([r(Y)r(Y)e,
.
J
are in
r(Z)z(Z)e],W) = -
Let
by term.
WE
J
is in
.
T
(r(Z)7(Z)e,
if
[r(Y)7(Y)e,r(Z)r(Z)e]
first term of
T
Then
=
Thus
that
Thus Lemma 2.2 implies the
E6 .
second and third terms of
Let
(v)
W E~
v(W)r(Y)r(Y)e)
by Lemma 2.1.
which proves that the
.
P . We shall consider
(J,w(W)e)
term
40.
([r(X)e,
= -
[v(y)r(Y)e,
v(Z)r(Z)e] ],
7(W)e)
(r([r(y)r(Y)e, v(Z)r(Z)e] )7(X)e, r(W)e)
[v(X)e,w(W)e])
= ([w(y)r(y)e,7(Z)r(Z)e],
= - ([r(y)v(Y)e,w(Z)r(Z)e],
[X,W])
(c)
= (r(Z)r(Z)e, r[X,W]w(Y)r(Y)e)
Now consider
([r(Z)r(Z)e,
[v(X)e,r(Y)v(Y)e] ],
7
= -
(v([v(X)e, v(Y)r(Y)e])7r(Z)r(Z)e,
= -
([v(X)e,7(Y)r(Y)e],
= (r(Y)r(Y)e, r([r(W)e,
[r(W)e,
(W)e)
w(W)e)
r(Z)7r(Z)e])
r(Z)r(Z)])v(X)e)
= (r(Y)r(Y)e, r(X)r(W)7(Z)r(Z)e)
(d)
Substituting (c) and (d) along with the equivalent expression for the third term gives
(J,r(W)e) = (r(Z)v(z)e, r[X,W]r(Y)v(Y)e)
+ (r(Y)w(Y)e,
r(X)r(W)r(Z)r(Z)e
-
r(X)v(W)r(Y)r(Y)e)
(v(Z)r(Z)e,
= 0
Case (vii).
u, v, w
in
T'
0
Essentially this is done by reducing it to cases
through (vi).
First consider
(i)
-L
a
[v(X)r(X)e,
[r(Y)r(Y)e, r(Z)w(Z)e]]
= [[X,w(X)e],
[v(Y)w(Y)e,
v(Z)r(Z)e]]
= [r(X)e, [[v(Y)w(Y)e, v(Z)T(Z)e],
+ [x, [w(X)e,
X]]
[I(Y)r(Y)e, w(Z)v(Z)e]]]
by case (ii)
= [r(X)e,
[r(Z)r(Z)e,
+ [I(X)e,
[X,w(Y)r(Y)e]]
[v(Y)v(Y)e,
[v(Z)r(Z)e,X]]
- [X,
[r(Z)v(Z)e,
[v(X)e, v(Y)v(Y)e]
- [X,
[v(Y)v(Y)e,
[I(Z)v(Z)e,
v(X)e]
by cases (iii) and
= [v(X)e,
[v(Z)v(Z)e,
- [r(X)e,
v(X)v(Y)v(Y)e]]
[v(Y)w(Y)e,
v(X)v(Z)v(Z)e]
- [X,
[v(Z)v(Z)e,
[v(X)e,
- IX,
[v(Y)v(Y)e,
[v(Z)v(Z)e,
v(Y)R(Y)e]
r(X)e]
Now consider
[w(Z)r(Z)e,
=
-
[w(Z)r(Z)e,
=
-
(Z)w(Z)e,
[v(X)w(X)e,
+ [w(Z)7(Z)e,
v(Y)r(Y)e]]
[v(Y)w(Y)e,
IX,
v(X)e]]]
[v(X)e, r(X)w(Y)r(Y)e]]
[X,
[r(X)e,
v(Y)v(Y)e]]
by case (iii)
]
(v)
r
42.
=
[v(Z)r(Z)e, v(X)e]]
[w(X)r(Y)w(Y)e,
+ [w(X)e,
[v(X)w(Y)r(Y)e,
+ [[v(X)e, 7(Y)r(Y)e],
r(Z)w(Z)e]]
r(X)w(Z)r(Z)e]
- [X, [[v(X)e, v(Y)v(Y)e]
, v(Z)v(Z)e]]
by cases (iv) and (ii)
If we write the corresponding expression for
[v(Y)r(Y)e,
[v(Z)r(Z)e, v(X)r(X)e]]
[w(X)w(X)e,
[v(Y)r(Y)e,
and combine we get
n(Z)v(Z)e]]
+ [w(Z)w(Z)e,
[r(X)r(X)e, v(Y)v(Y)e]]
+ [v(Y)r(Y)e,
[v(Z)v(Z)e,
r(X)r(X)e]]
= 0 .
So we have proven Jacobi Identity.
a Lie algebra.
Hence
.
is
Q.E.D.
We can now prove
Theorem 2.8.
Let
G/K
be a symmetric space and
w
an irreducible class-one orthogonal representation of
G
giving the imbedding
v: G/K ->
EN .
If the second
fundamental form of the imbedding is onto then
is a semi-simple non-compact Lie algebra with
; =
d =
e EN
e EN
a Cartan decomposition.
If
a
is the Cartan involution then in fact
is irreducible
orthogonal symmetric.
(Z,a)
43.
by
p
We define a representation
Proof.
p(G)I
P,(G) EN = V(G)
= AdG
u EE
Consider
action of u
and let
G
on
-
.
ad pu
be the adjoint
e .
on
Assert
ad-(p(g)u) = p(g)adiu
Let
be in
X
of
p(g)-1
f.
[w(g)u,XJ = -w(X)T(g)u
= -7(g)w(AdG
=
Let
v
be in
EN ,
X
p(g)oad uop(g - 1 )
(X)
in
([7(g)u,v],
[w(g)u,v]
So
-lX)u
= -
(v,
= -
(V(g-1)v,
=
([u, r(g- 1 )v],
=
(AdGg[u, r(g- 1 v],
w(X)w(g)u)
V (Ad g-X )u)
= p(g)oad uop(g -
AdG -lX)
x)
)
Assertion is thus proven.
Now let
is
G
B
= killing form on
.
Then by above
invariant hence is a constant multiple of the
Euclidean inner product on
constant is >0
Consider
ad e * ade
EN .
We now show that this
44.
r
adelk = 0
ade
is K-invariant.
e
Also let
= 0
I
since
W E I.
([e,v(X)7r(X)e],W)
but also
Then
I,
= -
Assert
(v(X)r(X)e,7(W)e)
= 0
if
W
is in
= 0
if
W
is in
by Lemma 1.4.
B;(e,e)
So to find
on
p
and
we need only consider
To
X E 6
Let
([e,[e,X]],X) = - ([e,X],[X,e])
(v(X)e,w(X)e)
(x,x)
tr(ade) 2
So
Let
X
and
Y
be in
.
([e,7(X)e],Y) = - (X,Y)
by above.
Hence
[e,r(X)e]
Hence
(ade)2
45.
([e,[e,w(X)e],w7(Y)e)
So
tr
= ( (X)e,7(Y)e)
(ade) 2 1T
0
B (e,e) = tr(ade) 2
But
tr(ade) 2 IT
T,o
= 2n .
is
We have thus shown that if
on
4
B g(X,X) > 0
,
X
EN .
is in
in
X is
If
if
the killing form
BZ(X,X)
= tr(ad X )2 + tr(v(X))
2
< 0
So
and
EN
being orthogonal under
, d. is semi-
Bo
simple.
Define
s:
-- > 4i
s(X+v)
Clearly
a linear map by
= X-v ,
s 2 = Identity.
X
in
Assert
v
s
in
EN
is a Lie algebra
automorphism.
s[X+v,Y+w]
= s[X,Y]
= [X,Y]
+ sv(x)w
- sv(Y)v + s[v,w]
- r(X)w + v(Y)v +
= [s(X+v),s(Y+w)]
4
v,w]
46.
So
s
is an involutive automorphism and
a Cartan decomposition.
The fact that
&=
EN
is
V. is irreducible
orthogonal symmetric is easily seen from the irreducibility
of the representation of
tion must be faithful.
on
Q.E.D.
EN
and the fact representa-
47.
Geometric Results.
§2.5.
We now apply the results we have obtained to the
problem of classifying those locally symmetric homogeneous spaces which have equivariant tight immersions.
We have the situation
7
G/K
locally symmetric ,
a real class-one representation of
immersion
r: G/K ->
we can assume
EN .
G
giving O-tight
By the corollary to Th. 2.3
is in fact irreducible and we get the
7
following classification theorems.
Theorem 2.9.
symmetric space and
Let
7
G/K
be an irreducible locally
an irreducible class-one
orthogonal representation of
7:
G/K -> EN .
(i)
v
(ii)
G/K
G
giving the immersion
Then the following are equivalent.
is 0-tight.
is a symmetric R-space and
7
is in fact
one of the imbeddings constructed in [12].
(iii) 7 is tight (has minimal total curvature).
Theorem 2.10.
space.
tion
Let
G/K
be a locally symmetric
Then the following are equivalent.
(i)
G/K
(ii)
There is an irreducible class-one representa-
7r
of
the immersion
G
covers a symmetric R-space.
such that the second fundamental form of
7: G/K -> EN
Proof of Theorem 2.9.
is
(i)
an onto map.
=> (ii).
-L
_-39~611
916--
48.
Since
r
G ,
is an irreducible representation of
r: G/K ->
the immersion
EN
is substantial and thus
since the immersion is 0-tight Th. 2.5 shows the second
is a semi-simple Lie algebra with
G
of
Thus
Int(e)
with Lie algebra L.
compact in
Int(;)
ding so
is the subgroup of
e
K
.
has eigenvalues
the compact subgroup
G
Th. 2.4 shows that
G
'e EN
;P =
onto; so Th. 2.8 shows
fundamental form is
is maximal
r
leaving
is an imbede
fixed, and
0, +1 ; so by definition
G/K
is a
symmetric R-space and the imbedding is one of the class
considered in [12].
(ii)
=> (iii).
(iii)
=> (i).
Shown in Th. 3.1,
See Th. 2.1, part (ii).
M'
is
the covering.
constructed in [12].
Then
=>
(ii).
Let
r
(i)
Proof of Theorem 2.10.
f: G/K ->
[12].
Suppose
be the imbedding
gives the required immer-
wof
sion.
(ii)
Int(;)
=> (i).
As above
4 = I
where
G
+ EN .
the isotropy subgroup of
is
maximal compact in
Thus
K
is a subgroup of
e , Ke , both are compact and
have the same Lie algebra.
Hence
G/K
covers
which is by definition a symmetric R-space.
Theorem 2,11.
Suppose
G/K
space which admits a motion group
is
L
Then
L
Q.E.D.
a locally symmetric
such that
simple, non-compact and properly contains
group.
G/Ke
G
L
is
as a sub-
is locally equivalent to an R-space.
49.
Proof.
If such a group
Theorem 3.1 a space
immersed in
G'/K'
ad(L)/ad(G)
equivalent to
.
[/72
exists then by
Hence
G'/K'
G/K
can be
covers an
Q.E.D.
R-space by Th. 2.11.
Remark.
L
We have confined our considerations to
irreducible spaces but the extension to reducible spaces
is easy as given by
Theorem 2.12.
If
G/K
is a locally symmetric space
which has a tight immersion then we can write
G/K = MlxM2 x.**Xh
where the
Proof.
Mi
Mi
are irreducible symmetric R-spaces.
We can write
G/K = MlXM2 x...xM
n
where the
are irreducible locally symmetric spaces.
G/K = M
Shall consider the case where
M2 .
The
general case follows easily.
The theorem follows from
Lemma 2.4.
immersions.
Let
f: M -> JRN
Then the immersion
fxg: MxM'
by
and
-> JRN+N
fxg (x,y) = (f(x),g(y))
g: M' ->
RN '
be
50.
is 0-tight if and only if
Proof of Lemma.
f
and
g
are 0-tight.
e: M -> JR1 ,
Let
: M' -> Ml
functions with non-degenerate critical points.
fn
p+X: MxM' -> JR
by
a critical point at
critical point of
If
X
is
e
and
if and only if
Y
x
is a
a critical point of index i and if
I+j
point of index-0 iff
so
ep,
cp+X
has
is a critical point of
critical point of index j then
point of index
Then the
(cp+*)(x,y) = c(x) + 4(y)
(x,y)
be
(x,y)
ay
is a critical
has only one critical
* have only one critical of
index-0, which gives Lemma and Theorem.
Q.E.D.
.
/s
cL
51.
CHAPTER 3.
EIGENVALUES OF THE LAPLACIAN.
1i. An Inequality on the Betti Numbers.
In this section we work with the formula given in
the remark following Def. 2.1.
ing
G/K
So we are tacitly assum-
is orientable throughout this section.
Definition 3.1.
If
f: M -> RN
is an immersion
then the absolute curvature at the point
m E M
is
defined by
Tm(f) =
Idet AJda
S
m
where
Sm
is unit sphere in
is volume of
.
We have if
For the immersions
EN)
T(G/K,,
Let
=V (G/K)
u E T0
Then
o
Y =
.
= (w,7(X)e)
endomorphism
of
)
.
= 0 .
since
r(g)e
X E
6
r(Y)7(g)e = v(g)r(X)e .
(7(g)u,w(Y)r(g)e)
Au
G/K -> 1RN
w(g)u E T
if and only if there is
Thus
r:
or )
N-1
Y E (G/K)g 0
(Ad g X)
V(G/K)
G/K
Lemma 3.1.
Proof.
M
with
Thus
Consider the
Then
(AuX,Y) = (u,w(X)r(Y)e)
52.
(A (g)uAd gX,Adg Y) = (u,r(X)w(Y)e) .
Thus
det Au = det AT(g)u
.
So
Q.E.D.
So to calculate the total absolute curvature we
need only calculate the absolute curvature at
0
and
this leads to
Theorem 3.1.
E bi(G/K, *) < V(G/K)
where
y
C2N-n-1
C2N-n-
n/2
n
is any eigenvalue of the Laplacian and
N
is
the complex dimension of any irreducible subspace of the
eigenfunctions with eigenvalue
sphere
Sr
y
.
Prof.
Consider once again the situation described
before Th. 1.3.
r: G -> VN
We have
representation and an immersion
Let
E To
and
X
and
A
Y E
E2 N
.
)
.
is a symmetric operator so we can choose
an orthonormal basis of
Let basis be
a class one unitary
v: G/K ->
(A X,Y) = (7r(X)r(Y)e,
Now
Cr = volume of
and
(Xi)
.
)
Then
such that
A
is diagonal.
53.
n
Det A
=
11 (V(Xi)v(Xi)e,)
i=n
Squaring both sides
2
jDet A g
= 1 (v(Xi),7(Xi)e,)
< H lv(X) )
2
2 n
E
arithmetic, geometric means
n(,) -
So
n
< (Y)n/2
< C2N-n- 1 (Y)n/2
T()
T(G/K,Tr,E
(Y)nn
-- n
IDet Ag
Thus
by Schwartz'
inequality
1v(Xi)
<
<
12
N)
V(G/K)
C2N-n-1
C2 N-1
Yln/2
The other inequality is merely a Morse inequality.
Remark.
Q.E.D.
Although this inequality is very weak it
would suggest that to find a tight imbedding the optimal
method would be to immerse in an irreducible subspace of
the eigenfunctions of the Laplacian and procede to
"eliminate unnecessary critical points."
ened by the next section.
This is strength-
3. Minimal Eigenvalues for R-Spaces.
We shall prove
Theorem 3.2.
For many symmetric R-spaces the tight
immersion is in fact an immersion in a space of eigenfunctions of the Laplacian with least eigenvalues.
(The term "many" will be clearer as we proceed.)
Something of this nature is done in [12].
But the
authors seemed to have ignored to a certain extent the
results of [18] especially that
X
nA-l
-
for any eigen-
value X of the Laplacian.
Proof.
We break the proof into a series of Lemmas.
First let us recall the idea of scalar curvature.
For convenience we shall define the scalar curvature at
a point
m E M
of a Riemannian manifold by
p(m) = -
E
(R(Xi,X )Xi,X.)
i,j
where
R
is the curvature tensor
orthonormal basis of
Lemma A.
of dimension
Let
n .
(Xi ]
is an
Mm .
G/K
be an irreducible symmetric space
Then
p(m) = n
Proof.
and
at all points.
See e.g. [12].
55.
Lemma B.
For the isometric immersion
7 G/K -> EN
(R(X,Y)Z,W) = (7(Y)7(Z)e,w(X)v(W)e)
-
Proof.
( (X) (Z)e,7(Y)7(W) e)
XY Z.)'W
This is of course just the classical Gauss
curvature equation but in our case has a very simple
proof.
(R(X,Y)Z,W) =
- ([[x,Y],z],w)
= -
(v[[X,Y],Z]e,w(W)e)
f[[X,Y],Z]e = 7(X)w(Y)r(Z)e
(R(X,Y)Z,W)
=
- V(Y)T(X)7(Z)e
(T(Y)v(z),w(X)v(w)e)
-
(v(X)r(Z)e,r(Y)v(W)e)
Q.E.D.
Now for
spaces we have
£=0
whe re
+E
N
EN = T o
T
Lemma C.
= (v(X)e)
The immersion
T
G/K -> EN
the sphere.
Proof.
Lemma D.
cf.
[12].
Let
Y, X E )
Then
lin.
hull (v(X)7(X)e]
is minimal in
56.
trace v(X)7(Y)IT
Proof.
this is
K
First let
=
H(X,Y) = trace v(X)w(Y)
H = kB .
invariant so
To
then one for
But
Now by Lemmas 1.4 and
2.2 we can choose an orthonormal basis
choosing one for
for some
constant K.
B(X,Y)
T
EN
0of
0
by first
so that we have
in matrix form
0A
A(X)
V(X)
(
(X)
-A
Thus
0
w(X)7(y) =
2 tr
A(X)A(Y)
=
2 tr
v(X)w(Y)
IT
0
Thus
tr
V (X)7w
(Y)T0 =kB
.Q.E.D.
be mean curvature normal at 0
If we now let
0o =
i
V (X
i
)
( X i ) e
Lemma E.
p(o) = (o0 , 0o) -
.
"-
I
57.
Proof.
p(O) =
i,j
-
=
R((X,XJ)Xi,X. )
J
(X
" (w(X )r(X)e(x)f
+ i ((X
(Xi )e,(X
i )
)e)
j ) T(X
j
)e)
(w(Xi ) w(Xi )e,w(X j) w(Xj )e) = ([o,)
Z
Now consider
(7r(X
i)
but
j
)e
,
fT(X )e)
E
w(X i ) v (X )e)
(v(X
r (X i ) r (X
i )
form orthonormal basis for
(v(X
)e,w(X
i ) 7(X
ij
Lemma F.
= -
(§ o,
)
= nX
i )v
(X
where
)e)
) e , r (X
To .
)e
Thus
=
-X
is an eigenvalue
of the Laplacian.
Proof.
By Lemma C
e(gK) = (v(g)e,e)
is
an
eigenfunction of the Laplacian and by Takahashi's Result
cited in Chapter
I-
n
e= -
(e,,e)
cfe
(e,e) = n/X
- Xce
say
I
58.
4
But
= E (v(Xi) v (X i )e,e)
APe()
=
So
(o,e)
= -
(e,e)
x(e,e)
,o = - Xe
( o,
= X2(e,e)
0)
= nX
Q.E.D.
Lemma G.
K+l
- n
Proof.
But
K
Direct from Lemmas A,
E
and
Q.E.D.
F.
can be calculated easily for many
spaces.
R
We shall do examples.
B.
is
G
invariant so
BZ(X,Y) = B
B,, = c B
(X,Y) + tr
= (l+k)B
The list
can be found in
.
But
v(X)w(Y)
.
[12] where credit is
given
to [8].
1.
Hermitian Symmetric Spaces.
By [91 the immersion is given as follows.
and
r
is equivalent to
AdG .
Thus
EN
_
tr 7 = tr ad X ad Y
59.
X, Y E
.
Thus
k = 1
X = 1
and
which is the
minimum value for eigenvalues of the Laplacian.
2.
Sphere.
Let
f: G/K -> Fy
Th. 1.3.
be the immersion constructed in
We can project onto a sphere radius 1 changing
the metric by a factor of
nn
curvature by
.
hence multiplying scalar
Y
The property of minimality is
unaltered
so by a result of Simons [20]
< n(n-1)
p(m) = nc
y -
but
with equality if and only
if G/K is the sphere
immersed in standard way
c = n
Y
This can also be deduced
from [18].
2.(n-1)
Thus for sphere immersed in
Y=
3.
standard way
n
2(n-l)
G/K = SO(2n)/U(n)
4 = SO(2n,
C)
EN = SO(2n)
l+k = 2
X = 1
which is
minimal by [18].
6o.
G/K = Sp(p+q)/Sp(p)
-
x Sp(q)
= su* (2 (p+q) )
1+k = 2 (p+q)
p+q1
p+q+l
minimal by [18].
Q.E.D.
61.
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BIOGRAPHY
64.
The author was born June 21st, 1945 in County Armagh,
Ireland.
Attended St. Macnissis College 1956-62.
from Queen's University,
Attended M.I.T.
Belfast in 1965.
1965-70.
Graduated