Some new obstructions to minimal and Lagrangian isometric immersions

Japan. J. Math. Vol. 26, No. 1, 2000

Some

new

obstructions

Lagrangian

isometric

to minimal

and

immersions

Dedicated to Professor Tadashi Nagano on the occasion of his seventieth birthday By

Bang-Yen

CHEN

(Received September 24, 1997) (Revised January 7, 1999) (from Kyushu Journal of Mathematics)

1.

Introduction

The main purpose of this paper is to introduce a new type of Riemannian curvature invariants and to show that these new invariants have interesting appli cations to severalareas of mathematics; in particular, they provide new obstructions to minimal and Lagrangian isometric immersions. Moreover, these new invariants enable us to introduce and to study the notion of ideal immersions. One of the most fundamental problems in the theory of submanifolds is the immersibility (or non-immersibility)of a Riemannian manifold in a Euclidean space (or more generally, in a space form). According to a well-knowntheorem of J.F. Nash, every Riemannian manifoldcan be isometrically immersed in some Euclidean spaces with sufficientlyhigh codimension. In order to study this fundamental problem, in view of Nash's theorem, it is natural to impose a suitable condition on the immersions. For instance, if one imposes the minimality condition on the immersions, it leads to PROBLEM 1. Given a Riemannian manifold M, what are the necessary con ditions for M to admit a minimal isometric immersion in a Euclidean m-space Em? It is well-knownthat for a minimal submanifold in Em, the Ricci tensor satis fies Ric_??_0.For many years this was the only known general necessaryRiemannian condition for a Riemannian manifold to admit a minimal isometric immersion in a Euclidean space. Themainresultsof this articlewerepresentedat the 3rd PacificRimGeometryConference heldat Seoul,Koreain December 1996;alsopresentedat the 922ndAMSmeetingheldat Detroit, Michiganin May1997.

106

BANG-YEN CHEN

An is

immersion

called

structure

J

space.

A

mov

real

of

real

[12]

states

of to

tangent

is

in

M

is

Cn

of

called

only

is

point

of

Hermitian if

its

almost

implies

normal M.

Cn,

a result

a Lagrangian

of

Gro

immersion TM_??_C

that

compact

M

complex

M=dimC

n-space

complexification

result

manifold

the

corresponding

if dimR

admits

if the

for

a

into

Euclidean M

Gromov's

a 3-manifold

M

Lagrangian

complex

immersions

Riemannian

of

n-manifold

trivial.

in

geometry)

space

in

if and

M

symplectic

tangent

a compact

Lagrangian

bundle From

each

immersion

that

bundle

n-manifold in

immersions

isometric)

obstruction

Riemannian isotropic

maps

totally Lagrangian

a

(or

M

For

necessary gent

of

totally

there

is

3-manifolds

in

(not

of no

the

tan

topological

C3,

because

the

trivial.

views,

it is

natural

to

ask

the

following

basic

ques

tion. PROBLEM manifold The A

class

when

do

author

in

exist

slant

slant

3.

to

The ƒÂ . He

admit

author

squared

mean

applying

his

been

(cf.

In

n1,

nk

...,

is

are

[3]

the

for

arbitrary

he

was

extensively paper

we

Riemannian are

extend

the

invariants

integers

satisfying

ask

a

satisfying

the few

invariant ƒÂ denoted

On

is

7].

immersions:

real

real.

[2,

admits

no

(or

A no

slant

Lagrangian

slant

the

for

a

immersions

other

hand,

there

in some equality

years a

this

space

solutions

for

denoted

invariant

Riemannian

to of

the

forms.

[3,

By

also

2

have

4, 8-11]).

n-manifold

We

the

1,

inequality

to

and ƒÂ(n1,...,nk),

2_??_n1_??_..._??_nk0

and

nature.

a personal

communication

(dated

April

(2.5) for

Riemannian

nian

space

forms.

3.

Sharp

We

recall

LEMMA

Then

n-manifolds

inequalities the

3.1.

2a1a2_??_ā,

following Let

with

the

equality

holding

only

involving ƒÂ(n1,...,nk) algebraic

a1,...,an,ā

equality

and

lemma be

holding

n+1

if

from real

and

only

[3]. numbers

if

such

a1+a2=a2=...=an.

that

for

Rieman

Obstructions

For positive

each

(n1,

to

...,

constants

minimal

and

Lagrangian

let

c(n1,

nk)•¸S(n),

given

isometric

...,

nk)

immersions

and

b(n1,

one

for

109

...,

nk)

denote

the

by

(3.1)

(3.2) The

following

theorem

THEOREM space

form

3.2.

is

Given

Rm(•¸)

of

the

an

constant

most

fundamental

n-dimensional sectional

submanifold

curvature •¸,

we

this

paper.

M

in

a

p •¸

M

if

Riemannian

have

(3.3) for

any

k-tuple

The there M

exists in

(n1,

equality an

Rm(•¸)

...,

nk)•¸S(n).

case

of

inequality

the

following

orthonormal

at p

take

basis

(3.3) e1,

...,

holds em

at

at p,

a point such

that

the

shape

and

only

if,

operators

of

forms:

(3.4)

where

I

is

an

identity

matrix

and

each

Arj

is

a

symmetric

nj•~nj

submatrix

that

(3.5) PROOF. Let

M

curvature

If be tensor

k=0,

this

a submanifold R

of

M

was of

satisfies

done

in

[5].

a Riemannian

Hence,

we

space

the

equation

of

and the squared

mean

curvature

assume

form

Rm(•¸).

k_??_1. The

Gauss:

(3.6)

The

scalar

(3.7)

curvature

H2 of M satisfy

Riemannian

such

110

BANG-YEN

where •ah•a2 Let

is

the

squared

(n1,...,nk)•¸S(n).

norm

of

the

CHEN

second

fundamental

form

h

of

M

in

Rm(•¸).

Put

(3.8) Then from (3.7) and (3.8) we find (3.9) Let L1, ..., Lk be mutually orthogonal subspaces of TpM with dim Lj=nj, j=1,...,k.

By choosing an orthonormal basis e1, ..., em at p such that

and en+1 is in the direction of the mean curvature vector, we obtain from (3.9) that (3.10) where

ai=hn+1ii,

i=1,...,n

and 'y = n + k

We put

Equation (3.10) is equivalent to (3.11)

where (3.12)

n3.

Obstructions

to minimal

and Lagrangian

isometric

immersions

111

Application of Lemma 3.1 to (3.11) yields (3.13)

On the other hand, (2.2) and Gauss' equation yield (3.14)

Combining (3.13) and (3.14), we get

(3.15)

where

0=Z (2.3),

If

equality

(3.15)

the are

(3.13),

Since the

exact

THEOREM

proof

of

proof

of

3.3.

any

k-tuple

p.

Let

quaternionic)

(n1,...,nk)•¸S(n).

a

by 3.2

be

Xpk).

(3.3).

point

p,

this

case,

(3.4)

verified

M

1)U...U(p

obtain

In

obtain

Theorem

(respectively,

(respectively,

at

Theorem

(3.16) for

at we

be

we

holds

equalities

can

xz

(3.15),

(3.3)

(3.15),

converse the

and in

and

Kaehlerian phic

,02=(z

(3.8)

actually

(3.14) The

tion,

U...Ut

From

and

then by

the

applying

based

also

an

n-dimensional

quaternionic) sectional

yields

space

Lemma

(3.13)

and

3.1,

(3.11),

Gauss'

equa

computation.

only

3.2

in

(3.5).

a straight-forward is

inequalities

on the

Lemma

curvature

and

following. totally

form

3.1

real

Mm(4•¸) 4•¸.

Then

of we

submanifold constant have

of holomor

a

112

BANG-YEN

4.

Some

Theorem

applications 3.2

THEOREM

gives

Let

admits

at

a

In

for

some

follows: totally

the

each

integer

vanishes immersion.

--*

Emj, of

_??_ 2 with

4.2.

scalar

If

any

Let

then

j

in =

and

to

there

Problem

exists

1.

a k-tuple

Riemannian

space

form

fri

the

the

>

be every

no

minimal

iso

(ni,...,n,)•¸S(n),

sharp.

This

minimal

can

be

submanifolds

n3)-space

product

Mn11,...,Mnkk Then

k

is

Riemannian

Clearly,

curvatures_??_0.

be

admits

codimension. k-tuple

4.1

, k,

a Euclidean

M of

each

Theorem

1, ...

of

identically.

COROLLARY

in

regardless

n_??_2

given

immersion

minimal

solution

that

immersion

space

invariant ƒÂ(n1,...,nk)

En-‡”nj

new

n-manifold.

such

Euclidean

For

:

geodesic

Then

following

(n1,...,nk)•¸S(n),

any

4.1.

f

p•¸M

isometric

on ƒÂ(n1,...,nk)

Let

the

a Riemannian

a point

k-tuple in

condition

to

if

immersion REMARK

the

be

minimal

particular,

point

metric

no

immediately M

and

M

Rm(•¸).

rise

4.1.

(ni,...,n,)•¸S(n)

then

CHEN

into

a Euclidean Mn11•~...•~Mnkk•~

immersion

f1•~...fk•~ƒÇ

manifolds

minimal

isometric

as

of

a

space.

product

Riemannian

seen and Ă

is

a

dimensions

immersion

(4.1) of

Mil

fi

xx

x

...

x

f k x

totally

scalar

vanishes

of

Since

identically.

such

sign way

of

k

En-E

the

ni

in

minimal

any

Euclidean

immersions

Mn11,...,Mnkk

curvatures_??_0,

equality

Mnjj

k x

m-space

f:

MJ

is

'

Emj,

a product j=

immersion 1,...,

k

and

a

immersion Ă.

PROOF.

in

t

geodesic

with

the

M~

the

Thus, of

that Riemannian

for

inequality enl+•••+nj+i, product,

are

Riemannian

invariant ƒÂ(n1,...,nk) any

minimal

(3.3)

holds

then

of Ml

immersion identically.

• •. ,

are the

second

manifolds

f

of 1

of Mn

Hence, tangent fundamental

k x E'E

in a Euclidean if we

to

dimensions_??_2

• x M~

the form

choose j-th h

space, e1,...,en

component of

f

satisfies

(4.2) for any Xj tangent to the j-th component of the Riemannian product according to Theorem 3.2. Now, Corollary 4.2 followsfrom Moore's lemma [14].

Obstructions REMARK 4.2.

which

satisfy The

Similar

result

holds

for isometric

immersions

113

immersions

H2=b(n1,...,nk)/c(n1,...,nk).

following

result

THEOREM damental

to minimal and Lagrangian isometric

4.3.

provides

Let

group ƒÎ1

or

(n1,...,nk)•¸S(n)

M

with

in

any

Lagrangian

isometric

be null

such

immersion

new a

solutions

compact

first

to

Riemannian

betti

n-torus

i.e., then

CTn

immersion

in

or

any

in

2 and

3.

n-manifold

number,

that ƒÂ(n1,...,nk)>0,

complex

Problems

M

Cn.

complex

with

b1=0.

If

admits

In

no

or

is

isometric

M

in

fun

a k-tuple

slant

particular,

n-torus

finite

there

admits

complex

no

Euclidean

n-space. PROOF. ƒ¿

=0,

If

then

this

M

case,

H is

nowhere

zero.

is

cannot

that not

the

class be

the

3.2

1-form

and ƒ¦

is

on

a

implies

that

defined

[ƒ¦] group

is

R)

is

assume ƒ¿•‚0. for

immersion

the

mean

[2]).

in

a

curvature

Thus, ƒ¦

compact

a non-trivial

H1(M;

a-slant

(cf. is

many

In

flat

vector

by ƒ¦(X)=n(cscƒ¿)•qH,JX•r.

1-form M

with

submanifold.

with ƒÂ(n1,...,nk)>0 a

M

closed

manifold

a minimal

we

admits

Because

Therefore,

cohomology

M

Theorem

[ƒ¦]•¸H1(M;R).

exact.

first

be

zero

Kaehlerian is

n-manifold

If

Let ƒ¦

it

Hence,

Riemannian

then

a flat

thus

a contradiction.

compact

M,

of and

and ƒ¦

cohomology non-trivial.

represents is

class Hence,

nowhere which

b1•‚0

a zero,

implies and

M

is

simply-connected.

with

covering

suppose finite map

M

is

lifts

Riemannian

universal

covering

Applying

the

a-slant

is

Theorem

COROLLARY damental

group ƒÎ1

some

mersion

damental

k-tuple in

also

4.4. or

compact.

we

of Let

with

may

complex

If

COROLLARY

4.5. or

Let with

M b1=0.

n-space be

to is

the

a

universal

a-slant

immersion

compact the

Riemannian

and ƒÎ1

covering

for is

map

finite,

preserves

the the the

a contradiction.

the

following

two

a compact If

Riemannian

6(ni,..., then

projective

with ƒÂ(n1,...,nk)>0

case,

results

by

using

some

4.3.

be a compact

b1=0.

M M

yields

prove

Theorem

M

this

Because

case

(n1,...,nk)•¸S(n),

the

group ƒÎ1

Since

previous

3.3, that

of

space.

the

to

n-manifold In

immersion

covering space

similar

Riemannian

group ƒÎ1•‚0.

invariant ƒÂ(n1,...,nk),

arguments

a compact

fundamental

universal

for

submanifold

submanifold

yields

a

nowhere

Now, and

is

n-manifold

cohomology ƒ¦

a-slant

(n1,...,nk)•¸S(n).

Kaehlerian

Then ƒ¦

an

3.2

M

k-tuple

is

a complex

Theorem

Suppose some

M is

o(ni,...

n-manifold

nk)>2 M

admits

(n(n no

with

-1)

finite n3(nj

Lagrangian

isometric

fun -1)) im

CPn(4). Riemannian , nk)>2(~

n-manifold n3 (n3

with -1)

finite n(n

fun -1))

114

for

BANG-YEN

some

k-tuple

mersion

in REMARK

laries

4.4

(n1,...,nk)•¸S(n),

the

complex 4.3.

and

w:Sn•¨Cn

are

then

hyperbolic The

4.5

M

n-space

conditions

sharp.

defined

CHEN

For

admits

no

Lagrangian

isometric

im

CHn(-4).

on ƒÂ(n1,...,nk) example,

given

consider

in

Whitney's

Theorem

4.3,

Lagrangian

Corol

immersion

by

(4.3) with

y20+y21+...+y2n=1.

point

The

Whitney's

immersion

has

a unique

self-intersection

w(-1,0,...,0)=w(1,0,...,0). For

any

respect

to

n-tuple

the

(n1,...,nk)•¸S(n),

induced

metric;

we

have ƒÂ(n1,...,nk)_??_0

and ƒÂ(n1,...,nk)=0

only

on

at

the

Sn

unique

with

point

of

self-intersection. REMARK ometric curvature infra

of

face

in

Let and

p)

M =: 1•~Sn-1

each

must

let Ă:

=

F(s) 51

x to

4.3

5. In

A general other

MAXIMUM clidean

by

Lagrangian

submanifold

M

over

all

at unit

some

®

t(p),

Sn-1 a point

p

e

into Cn in Cn

x

Sn^1.

m-space

there

does

not

have

If

a k-tuple

(5.2)

(n1,...,nk)•¸S(n),

is

of

Cn M.

an

and

that

both

the

and

ideal

For

satisfy

Lagrangian sur

M. plane

given in

complex

isometric pair

must

Ricci

Lagrangian

hypersphere the

ge

the

by

En

at

defined

by

Lagrangian

immersion

{(u,p),(-u,-p)}

of and on

conditions

centered

extensor

b1(M)=1,

F(s)=

points

in S

moreover, M

on ƒÎ1(M)

is

a

and

for

positive b1(M)

in

removed.

PRINCIPLE. Em.

unit

each

in

compact

complex

Cn

f

carries

principle

we

the

. Clearly, ƒÎ1(M)=Z

shows be

maximum

hand,

the

--~

Then

which

on

following

namely,

the ƒÂ-invariant ƒÂ(n1,...,nk)

example cannot

Sn-1

in

every

is the

Cn;

vectors of

points

circle be

: 51

tangent

curvature

exist the Let

it

a direct

immersions

relationship

between

the

new

invariants.

following.

satisfies

M the

(5.1) for

unit

Gaussian

the

f

4.3 in

nonpositive be

Theorem

submanifolds

(n1,...,nk)•¸S(n), This

Theorem

runs the

of

Lagrangian

Sn-1•¨En(n_??_3)

k-tuple

the

be

Denote

constant.

On

u that

S1•¨C

origin.

f (s, of

where

F:

consequence

compact

means

C2

immediate

compact

every

this

M

the

An on

Ric(u)_??_0,

surfaces,

eis

4.4.

condition

then

be

an

equality

n-dimensional case

of

submanifold (3.3),

i.e.,

it

of satisfies

a

Eu

Obstructions

for

any

to minimal

(m1,...,mj)•¸S(n),

and Lagrangian

isometric

immersions

115

where

(5.3) PROOF. For Em,

It

an

follows

from

isometric

Theorem

3.2

Theorem

immersion

x:

3.2

with •¸=0.

M•¨Em

of

a

Riemannian

n-manifold

M

in

yields

(5.4) where Ģ0

is

the

Riemannian

invariant

on

M

defined

by

(5.5) Inequality

(5.4)

enables

us

to

introduce

the

notion

of

ideal

immersions

as

follows. DEFINITION Em

is

called

5.1. an

REMARK an

isometric

the

least

at

point

curvature a

5.1

on

Riemannian

the

each

an

k-tuple

on

receives

REMARK

5.2.

is

the

from

The

isometric

In

this

Laplacian Ģ

we

above

the

it

is

principle

an

yields

M

surrounding

thus

space the

mean

the

of

squared

amount

mean of tension

point. the

Laplacian

following

important

(5.1)

and between

the

that

receives

immersion

the that

fact M

that

isometric

in

for

a

given

automatically.

relationship and

the

equality

immersion

of

close

manifold

at

satisfies ideal

The

fact

an

M

identically.

that

measures

space

eigenvalues some

for

manifold;

M•¨Em

(5.4)

means

well-known

field

surrounding

x:

from

simply

maximum

establish

Riemannian

and

tension

of

Immersions).

by Ģ0(p))

(5.4)

n-manifold

case

immersion

Riemannian

the

Riemannian

equality

Ideal

ideal

submanifold

between

a

to the

then

section on

due

immersion

Relations

an

(given

another

(n1,...,nk)•¸S(n),

6.

tension

exactly in

point

submanifold

If

This is

manifold at

fact:

M. field

is

a

the of

M•¨Em of

of

satisfies

Interpretation

x:

amount

p

immersion if it

(Physical

vector

curvature

isometric

immersion

immersion possible

each

An

ideal

new

the

invariants

eigenvalues

invariants.

In

of

the

particular,

we improve a well-knownresult of T. Nagano obtained in [15]. In order to do so, we recall (cf.

[6]

for

Let M.

the notions

Denote

of order

and type

from the theory

of finite

type

submanifolds

details). M

be

a compact

by ăj

the

Riemannian j-th

nonzero

n-manifold eigenvalue

and Ģ of

the

the

Laplacian Ģ

Laplacian on

operator M.

of

116

BANG-YEN

For

an

isometric

immersion

x:

CHEN

M•¨Em

of

M

in

Em,

let

(6.1) denote

the spectral

decomposition

of x, where

x0 is center

of mass

of M in Em.

The set

(6.2) is called the order of the submanifold. The smallest element p in T(x) is called the lowest order of x and the supremum q of T(x) is called the highest order of x. The immersion is said to be of finite type if the highest order q is finite; and it is said to be of infinite type if the highest order q is infinite. Moreover, the immersion is said to be of k-type if T(x) contains exactly k elements. Clearly, the immersion is of 1-type if and only if p=q. In this case, the immersion is called a 1-type immersion of order {p}. For an isometric immersion of a compact Riemannian manifold M in a Eu clidean space, one has (cf. [6]) (6.3) where vol(M) denotes the volume of M and p, q are the highest order and the lowest order of M, respectively. Either equality sign of (6.3) holds if and only if the immersion is of 1-type with order {q} or order {p}, respectively. Since p_??_1,(6.3) implies immediately the followingresult [17]: (6.4)

THEOREM Riemannian

6.1.

Let

n-manifold.

x:

M•¨Em

be

an

isometric

immersion

of

a

compact

Then

(6.5) for

any

k-tuple

The of

order

(n1,...,nk)•¸S(n),

equality {q}

satisfying PROOF.

case

associated

of with

H2=Ģ0(n1,...,nk) Follows

where (6.5)

holds

if

(n1,...,nk),

q and

is

the

only

i.e.,

x

is

and

(6.3).

identically. from

Theorem

3.2

highest if

x

is

a 1-type

order a

of

1-type immersion

the

immersion.

ideal

immersion of

order

{q}

Obstructionsto minimal and Lagrangianisometricimmersions

117

Theorem 6.1 implies immediately the following. THEOREM6.2. If a compact Riemannian n-manifold M admits a 1-type iso metric immersion in a Euclidean space, then

(6.6) for

any

k-tuple

The if

the

of

an

in

terms

the

(n1,...,nk)•¸S(n),

equality

case

1-type

immersion

REMARK

6.1.

isometric

where

of

(6.6) is

of

our

an

ideal

Theorem

immersion new

Riemannian

holds

6.1 of

of

provides

is

the

order

k-tuple

associated us

a

1-type

immersion. if

to

and

only

(n1,...,nk).

estimate

the

manifold let

with

the

with

way

example,

2-spheres

of

(n1,...,nk)•¸S(n)

Riemannian

For two

a

immersion

a compact

invariants.

product

p

for

in

highest

order

a Euclidean

space

M=S2(1/a2)•~S2(1/b2)

curvatures

1/a2,

be 1/b2,

respectively.

Assume

for of

some M

and

natural

are

number

given

by

s•¸N0.

2/a2,

4Ģ0=1/(a2b2)>ăs,

isometric

q of

must

be The

of

any

first

of

isometric

the

s nonzero

6.1

implies in

of

Since Ģ0=1/(4a2b2)

that

any

codimension,

immersion

eigenvalues ă1, ă2,...,ăs

respectively.

S2(1/a2)•~S2(1/b2)

independent

order

the

Theorem

immersion

Consequently,

Then

6/a2,...,s(s+1)/a2,

the

highest

Euclidean

if b is

order

space

a small

S2(1/a2)•~S2(1/b2)

is

number, in

any

at

of

least the

any s+1.

highest

Euclidean

space

large. following

result

provides

a

sharp

relationship

between ă1

and

our

new

invariants. THEOREM irreducible M

isotropy

6.3.

If

M

action,

is then

a compact the

homogeneous

first

nonzero

Riemannian eigenvalue ă1

n-manifold of

the

Laplacian

with on

satisfies

(6.7) for

any

k-tuple

(n1,...,nk)•¸S(n).

Therefore,

we

have

(6.8) The equality sign of (6.8) holds if and only if M admits a 1-type ideal immer sion in a Euclidean space.

118

BANG-YEN

PROOF. neous ric

Follows

Riemannian immersion

of

Riemannian

order

REMARK [15],

In

general,

we

Theorem

and

inequality

6.4. M

k-tuple

yields

the to

Let

admits

where ƒÏ and ƒ¢0>ƒÏ

manifold

THEOREM

M an

(n1,...,nk),

the

fact

isotropy

that

every

action

compact

admits

of

homoge

a 1-type

a compact

isomet

homogeneous

constants.

have ƒ¢0_??_ƒÏ, also

and

invariants ƒÂ(n1,...,nk)

When k=0,

3.2

6.2

irreducible

namely, ƒÉ1_??_nƒÏ,

Riemannian

If

are

6.2.

Nagano

(1)

Theorem with

{1}

manifold

T.

pact

from

manifold

CHEN

ideal

for

a

ideal compact

the

reduces

to

a well-known

normalized

most

following

admit be

(6.7) is

scalar

Riemannian

necessary

result

curvature

of

of

M.

manifolds.

intrinsic

condition

for

a

com

immersions. Riemannian

immersion

in

n-manifold.

a Euclidean

space

associated

with

a

then

(6.9)

(2) If M satisfiesap

1),

(n and

>

A1=

admits

M

nL1(ni,...

8(n =

6.6

RPn(1); +

L~(2,

an

L 0

the

only

=

only

in

a

n-dimensional

if and

the =

following

4(n+

1),

o(n,

n,

2, 2, 2, 2, 2, 2, 2)

is

scalar

and Al

1),

immersion

positive

, nk)

=

CP1

ideal

For

for

Ao

6.8.

with

Theorem

=1

Al

48,

an

PROOF. ifold

Leo

1);

COROLLARY which

from

1),

n, n)

irreducible

0(n, +

for

have

n)

=

3)/n

(n+3)/n

for

HP(4)

0P2(4).

Hermitian

symmetric

arbitrary

homogeneous we

(n for

space

compact

=

=

=145/56

Euclidean

curvature,

facts: zo

Einstein

nă1=4Ą

space

codimension. Kaehler

[16].

man

Thus,

ai>

if

(6.12) The

coefficient

three

cases

of ƒÂ(n1,...,nk) occurs:

n1+n2=n.

In

compact

type,

irreducible

the

namely,

7.

equality

we case

EXAMPLE is

M

a

7.1.

a

is an

is

equal

(2)

k=1,

compact

to

type

with

some

imply

examples

integer

k•¸{0,1,...,[n/2]},

and

only

that

if

and

one (3)

symmetric geometric

(6.12)

of

the

k=2, space

of

properties

holds

unless

of n=2,

line.

satisfying

simple

if

Hermitian

projective

submanifolds

two n=n1+1,

irreducible

together

complex

some

inequality

hypersurface

of

is

of

provide of

M

observations

spaces

Examples

Here,

when

above

unless

(6.12)

k=0, n=2,

particular,

symmetric

k=0;

Sn-k

(1)

in

basic of

equality

submanifolds

which

satisfy

the

(3.3).

For

any of

En+1

satisfying

(5.1)

a spherical with

(n1,...,nk)=(2,...,2).

cylinder

Ek•~

120

BANG-YEN

EXAMPLE in

Rm(•¸)

lar,

a

7.2.

For

satisfies

the

horosphere

each

equality

in

a

in

(3.3)

EXAMPLE round

a2+b2=1,

is an

ideal

with

7.3.

k-sphere

with

in

(3.3)

n=2k, with

space

A

equality

number

hyperbolic

(n1,...,nk)=(2,...,2). the

even

CHEN

totally

a totally

umbilical

the

submanifold

be

curvature

product

embedding

embedding

In

satisfies

equality of

in

Rm(•¸)

particu (3.3)

also

with

satisfies

k=0.

constant

the

submanifold

(n1,...,nk)=(2,...,2).

H2k+1

Let ƒÇa:Sk(1/a2)•¨Ek+1

with

umbilical

in

E2k+2

which

1/a2.

For

satisfies

the

the

standard

any

two

equality

embedding

positive

in

(3.3)

numbers

with

of

a

a,

b

(n1,...,nk)

=(2,...,2). In some

general,

if l

integers

and

n-l

k>1,t,s,

Sl(1/a2)•~Sn-l(1/b2)

in

(n1,...,nk),

where

EXAMPLE ative

For

NB2k

austere

lies

(3.3)

with

in

M to

be

THEOREM

(8.1)

an

ideal

such

that l=kt, of

embedding

associated

austere we

the

n-l=ks

Riemannian

for product

with

the

k-tuple

and open

the

dense

with

submanifold

=1

and

identity

Sn+1(1)

(e,TPB2JC)

map

subset

relative

of

with

zero

rel

put

I (,)

inequalities

a submanifold we

put

U

nullity

Mm(4•¸)

for

from

of

n-2k.

=

NB2k

NB2k.

0}.

into

Moreover, NB2k

of

of JX, of result

Mm(4•¸) 8.1. with •¸