Comparison Theorems in Riemannian Geometry

J.-H.Eschenburg

0.Introduction

The subject of these lecture notes is comparison theory in Riemannian geometry:

What can be said about a complete Riemannian manifold when (mainly lower) bounds

for the sectional or Ricci curvature are given?Starting from the comparison theory for

the Riccati ODE which describes the evolution of the principal curvatures of equidis-

tant hypersurfaces,we discuss the global estimates for volume and length given by

Bishop-Gromov and Toponogov.An application is Gromov's estimate of the number

of generators of the fundamental group and the Betti numbers when lower curvature

bounds are given.Using convexity arguments,we prove the"soul theorem"of Cheeger

and Gromoll and the sphere theorem of Berger and Klingenberg for nonnegative cur-

vature.If lower Ricci curvature bounds are given we exploit subharmonicity instead

of convexity and show the rigidity theorems of Myers-Cheng and the splitting theorem

of Cheeger and Gromoll.The Bishop-Gromov inequality shows polynomial growth of

nitely generated subgroups of the fundamental group of a space with nonnegative Ricci

curvature (Milnor).We also discuss brie y Bochner's method.

The leading principle of the whole exposition is the use of convexity methods.

Five ideas make these methods work:The comparison theory for the Riccati ODE,

which probably goes back to L.Green [15] and which was used more systematically by

Gromov [20],the triangle inequality for the Riemannian distance,the method of support

function by Greene and Wu [16],[17],[34],the maximum principle of E.Hopf,generalized

by E.Calabi [23],[4],and the idea of critical points of the distance function which was

rst used by Grove and Shiohama [21].We have tried to present the ideas completely

without being too technical.

These notes are based on a course which I gave at the University of Trento in March

1994.It is a pleasure to thank Elisabetta Ossanna and Stefano Bonaccorsi who have

worked out and typed part of these lectures.We also thank Evi Samiou and Robert

Bock for many valuable corrections.

Augsburg,September 1994

J.-H.Eschenburg

1

1.Covariant derivative and curvature.

Notation:By M we always denote a smooth manifold of dimension n.For p 2 M,

the tangent space at p is denoted by T

p

M,and TM denotes the tangent bundle.If

M

0

is another manifold and f:M!M

0

a smooth (i.e.C

1

) map,its dierential at

some point p 2 M is always denoted by df

p

:T

p

M!T

f(p)

M

0

.For v 2 T

p

M we write

df

p

(v) = df

p

:v = @

v

f.If c:I!M is a (smooth) curve,we denote its tangent vector

by c

0

(t) = dc(t)=dt = dc

t

:1 2 T

c(t)

M (where 1 2 T

t

I = IR).If f:M!IR,then

df

p

2 (T

p

M)

.If M is a Riemannian manifold,i.e.there exists a scalar product <;>

on any tangent space of M,this gives an isomorphism between T

p

M and (T

p

M)

;the

vector rf(p) corresponding to df

p

is called the gradient of f.

Let M be a Riemannian manifold.We denote by <;> the scalar product on M

and we dene the norm of a vector by

kvk =

p

< v;v >;

the length of a curve c:I!M by

L(c) =

Z

I

kc

0

(t)kdt;

and the distance between x;y 2 M by

jx;yj = inf fL(c);c:x!yg:

where c:x!y means that c:[a;b]!M with c(a) = x and c(b) = y.If L(c) = jx;yj

for some c:x!y,then c is called shortest.The open and closed metric balls are

denoted by B

r

(p) and D

r

(p),i.e.

B

r

(p) = fx 2 M;jx;pj < rg;D

r

(p) = fx 2 M;jx;pj rg:

Similarly,we dene B

r

(A) for any closed subset A M.

We denote by X(M) the set of vector elds on M.

Denition 1.1 The Levi-Civita covariant derivative

D:X(M) X(M)!X(M)

(X;Y )!D

X

Y;

is determined by the following properties holding for all functions f;g 2 C

1

(M) and

for all vector elds X;X

0

;Y;Y

0

2 X(M):

2

1.D

(fX+gX

0

)

Y = fD

X

Y +gD

X

0

Y;

2.D

X

(fY +gY

0

) = (@

X

f)Y +fD

X

Y +(@

X

g)Y

0

+gD

X

Y

0

;

3.D

X

Y D

Y

X = [X;Y ] ="Lie bracket";

4.@

Z

< X;Y >=< D

Z

X;Y > + < X;D

Z

Y >.

Denition 1.2 The Riemannian curvature tensor (X;Y;Z) 7!R(X;Y )Z is dened as

follows:

R(X;Y )Z = D

X

D

Y

Z D

Y

D

X

Z D

[X;Y ]

Z

It satises certain algebraic identities ("curvature identities"),namely

< R(X;Y )Z;W >= < R(Y;X)Z;W >= < R(X;Y )W;Z >=< R(Z;W)X;Y >

and the Bianchi identity

R(X;Y )Z +R(Y;Z)X+R(Z;X)Y = 0

(cf.[29]).In particular,

R

V

:= R(:;V )V

is a self adjoint endomorphism of TM for any vector eld V on M.Several notions of

curvature are derived from this tensor:

1.Sectional curvature K(;):For every linearly independent pair of vectors X;Y 2

T

p

M,

K(X;Y ) =

< R(X;Y )Y;X >

kXk

2

kY k

2

< X;Y >

2

:

K is dened on the space of two dimensional linear subspaces of T

p

M (depending

only on span(X;Y )).

2.Ricci curvature

Ric(X;Y ) = trace(Z 7!R(Z;X)Y):

By the curvature identities,Ric(X;Y ) = Ric(Y;X).Ricci curvature in direction

X is given by

Ric(X) = Ric(X;X)

where X is a unit vector.

3.Scalar curvature

s = trace(Ric) =

X

Ric(E

i

;E

i

)

3

where fE

i

g

ni=1

is a local orthonormal basis.

There is a close relationship between R

V

= R(:;V )V and the sectional curvature:

Let kV k = 1.For X orthogonal to V we have

< R

V

X;X >=< R(X;V )V;X >= K(V;X)kXk

2

Hence the highest ("

+

") and lowest ("

") eigenvalues of R

V

give a bound to K(V;X),

since

(R

V

)

< R

V

X;X >

< X;X >

+

(R

V

):

Moreover,trace(R

V

) = Ric(V;V ).

Let us come back to the covariant derivative.It is easy to see that for any p 2 M,

(D

X

Y )

p

depends only on dY

p

:X(p) where the vector eld Y is considered as a smooth

map Y:M!TM.Therefore,the covariant derivative is also dened if the vector

elds X and Y are only partially dened.E.g.if :I!M is a smooth regular curve

and Y is a vector eld along ,i.e.a smooth map Y:I!TM with

Y (t) 2 T

(t)

M

for all t 2 I (e.g.

0

is such a vector eld),then

Y

0

(t):=

DY (t)

dt

:= D

0

(t)

Y

is dened (just extend

0

and Y arbitrarly outside ).Similar,if :I

1

:::I

k

!M

depends on k variables,we have k partial derivatives

@

@t

i

and corresponding covariant

derivatives

D

@t

i

(i = 1;:::;k) along .(Formally,a vector eld along is a section of the

pull-back bundle

TM,and D induces a covariant derivative on this bundle.)

Denition 1.3 A vector eld Y along a curve :I!M is called parallel if Y

0

= 0.A

curve is a called a geodesic in M if

0

is parallel,i.e.if

(

0

)

0

= D

0

0

= 0:(1:1)

(1.1) is a 2

nd

order ODE.In fact,if x = (x

1

;:::;x

n

):M!IR

n

is a coordinate

chart with E

i

=

@

@x

i

and if we put

D

E

i

E

j

=

kij

E

k

4

(summation convention!),then

0

= (

i

)

0

E

i

where

i

:= x

i

,and

D

dt

0

= (

i

)

00

E

i

+(

j

)

0

D

0

E

j

with

D

0 E

j

=

k

D

E

k

E

j

= (

k

)

0

ikj

E

i

;

hence (1.1) is equivalent to

(

i

)

00

+(

j

)

0

(

k

)

0

ikj

= 0

To some extent,Riemannian geometry is the theory of this ODE.

Denition 1.4 For any v 2 TM let

v

denote the unique geodesic with

0

(0) = v.For

s;t 2 IR with jsj and jtj small we have

sv

(t) =

v

(st) by uniqueness for ODE's.Thus

for v 2 TM with kvk small enough,

exp(v):=

v

(1)

is dened and gives a smooth map exp:(TM)

0

!M where (TM)

0

is a neighborhood

of the zero section of TM.This is called the exponential map of M.M is called

(geodesically) complete if exp is dened on all of TM.Fixing p 2 M,we put exp

p

=

expjT

p

M.

Remark 1.5 The map exp

p

is a dieomorphism near the origin (in fact,d(exp

p

)

0

is

the identity on T

p

M),and it maps all the lines through the origin of T

p

M onto the

geodesics through the point p 2 M.Thus,exp

p

:T

p

M!M can serve a a coordinate

map near p ("exponential coordinates") which preserves the covariant derivative at p,i.e.

covariant dierentiation at p is the same as taking the ordinary derivative at 0 2 T

p

M in

exponential coordinates.To see this,identify M and T

p

M near p via exp

p

and consider

the ordinary derivative D

o

= @ on T

p

M.It satises the rules (1.),(2.) and (3.) of the

Levi-Civita derivative (but not 4.for the Riemannian metric).Hence the dierence

= D@ is a tensor eld,i.e.

X

fY = f

X

Y for all functions f (by 2.),further it is

symmetric,i.e.

X

Y =

Y

X (by 3.),and it satises

v

v = 0 for all v 2 T

p

M,since D

and @ have the same geodesics at p.Thus j

p

= 0.

5

2.Jacobi and Riccati equations;equidistant hypersurfaces.

Equation (1.1) is a nonlinear ODE which in general cannot be solved explicitly.

Therefore,we consider its linearization.This is the ODE satised by a variation of

solutions of (1.1),i.e.of geodesics.So let (s;t) =

s

(t) be a smooth one-parameter

family of geodesics

s

.Put V =

@

@t

2 X(

s

) and J =

@

@s

.Then J is the variation vector

eld and V the tangent eld of the geodesics

s

,hence D

V

V = O.

Fig.1.

Then we have

J

00

=

D

@t

D

@t

J =

D

@t

D

@t

@

@s

:

We can interchange the order of dierentiation,getting

J

00

=

D

@s

D

@t

@

@t

+R(V;J)V;

J

00

+R(J;V )V = 0:(2:1)

Equation (2:1) is called Jacobi equation.

Denition 2.1 A vector eld J along a geodesic is called a Jacobi eld if it satises

the Jacobi equation.

Remark 2.2 J is a Jacobi eld along if and only if

J(t) =

d

ds

0

s

(t) (2:2)

6

for some one-parameter family of geodesics

s

with

0

= .

Implication"("was shown above.To prove the opposite implication,we have to

construct the family

s

.Let (s) = exp

(0)

sJ(0).Let X be a vector eld along such

that X(0) =

0

(0) and X

0

(0) = J

0

(0) and put

s

(t) = exp

(s)

tX(s) (2:3):

If we put

~

J =

@

@s

0

s

;

then,by"(",

~

J satises the Jacobi equation.Since

~

J(0) = J(0) and

~

J

0

(0) =

D

@t

@

@s

j

(0;0)

=

D

@s

@

@t

j

(0;0)

=

D

@s

X(s)j

0

= X

0

(0) = J

0

(0);

we get J =

~

J by uniqueness of the solution.

Next,we want to split this 2

nd

degree equation in a system of 1

st

degree equations.

To do this,we embed the 1-parameter family of geodesics describing the Jacobi eld

into an (n 1)-parameter family.I.e.we choose a suitable smooth map

:S I!M

where S is an (n 1)-dimensional manifold,such that

s

(t) = (s;t) is a geodesic for

any s 2 S.If is a regular map,then V = d (

@

@t

) can be viewed as a vector eld on an

open subset of M with D

V

V = 0,and the Jacobi elds J arising from variations in S-

directions commute with V,i.e.we have [J;V ] = 0 or

D

V

J = A J (2:4)

where A = DV,i.e.A X = D

X

V.This is the rst of our equations:Knowing A,we

recieve J by solving a 1

st

order equation.

It remains to derive an equation for A.Let us consider rst an arbitrary vector

eld V on M and let A = DV as before.In general,the covariant derivative of a tensor

eld A is dened by

(D

V

A)X = D

V

(AX) A D

V

(X):

7

Hence we have

(D

V

A)X = D

V

D

X

V A(D

X

V +[V;X])

= D

X

D

V

V +R(V;X)V +D

[V;X]

V A

2

X A[V;X]

= D

X

D

V

V +R

V

(X) A

2

X

:

Therefore

D

V

A+A

2

+R

V

= D(D

V

V ):(2:5)

If we suppose D

V

V = 0 (i.e.the integral curves

s

are geodesic),then we get an ODE

for A,the so called Riccati equation

A

0

+A

2

+R

V

= 0:

Thus we have split the Jacobi equation J

00

= R

V

J in two equations as follows:

J

0

= AJ (2:6)

A

0

+A

2

+R

V

= 0:(2:7)

We note that the second equation can be solved independently of the rst.

Let us consider now the important special case where (DV )

= DV,that is

< D

X

V;Y >=< X;D

Y

V >

for all vector elds X,Y.Then V is locally a gradient,i.e.locally V = rf for some

function f:M!IR.Consequently,< V;V > is constant,since

@

X

< V;V >= 2 < D

X

V;V >= 2 < D

V

V;X >= 0:

Thus we may assume that < V;V >= 1.Now let us consider the level hypersurfaces

S

t

= fx 2 M:f(x) = tg:

Since V = rf 6= 0,the S

t

are regular hypersurfaces and V j

S

t

is the unit normal vector

eld on S

t

.Thus in this case,our (n1)-parameter family of geodesics :S I!M

is given by

(s;t) = exp(t t

0

)V (s)

8

where S = S

t

0

for some t

0

2 I,and f( (s;t)) = t,or in other words,S

t

=

t

(S) where

t

(s):= (s;t).Such a family of hypersurfaces S

t

is called equidistant,and the function

f t

0

is called the signed distance function of the hypersurface S = S

t

0

.In fact we have

jf(x) t

0

j = jx;Sj:= inf

s2S

jx;sj (2:8)

for x in a small neighborhood of S.Namely,if c:[a;b]! (S I) M is a curve with

c(a) 2 S

t

0

and c(b) 2 S

t

1

,then we have c(u) = (s(u);t(u)) with t(a) = t

0

,t(b) = t

1

,

and

kc

0

(u)k

2

= kd :s

0

(u)k

2

+t

0

(u)

2

t

0

(u)

2

;

hence its length is

L(c)

Z

b

a

jt

0

(u)jdu jt(b) t(a)j jt

0

t

1

j:

Fig.2.

In this case,all the quantities discussed above have geometric meanings.The

Jacobi elds J(t) = d

(s;t)

(x;0) = d

t

x for x 2 T

s

S measure the change of the metric of

S

t

=

t

(S) when t is changed;in fact,kJ(t)k=kJ(t

0

)k is the length distortion between

S

t

and S.Moreover,A = DV,restricted to the hypersurface S

t

,is the shape operator

of S

t

since V jS

t

is a unit normal vector eld on S

t

.Its eigenvalues are called principal

curvatures,their average the mean curvature of S

t

.Since Equation (2.7) is nonlinear,

A(s;t) can develop singularities which are called focal points of S.Let us see some

examples.Example 2.3 Let S

t

= @B

t

(p),where B

t

(p) = fx 2 M:jx;pj < tg is the Riemannian

ball.

9

Then V is radial and

A(t)

1

t

I as t!0 (2:9)

because a Riemannian manifold behaves as a Euclidian space for t!0.

Example 2.4 ([9]) Let us suppose that R

V

= kI,k 2 IR,that is M has constant

curvature.Moreover let us suppose that A = aI,where a is a real function dened on

M (A is the second fundamental form of a family of umbilical hypersurfaces).In this

case equation (2.7) becomes:

a

0

+a

2

+k = 0:

If k > 0,then M is a sphere (if it is assumed to be complete and simply connected).

The solutions are given by

a(t) =

p

k cot(

p

k(t t

0

)):

This corresponds to the fact that there is (up to congruence) only one equidistant

family of umbilical hypersurfaces in the sphere,namlely concentric Riemannian spheres

(latitude circles).

Fig.3.

If k = 0,then M is a Euclidean space and the solutions are

a(t) =

1

(t t

0

)

;a(t) = 0:

10

Fig.4.

These solutions correspond to the three umbilical parallel hypersurface families in eu-

clidean space:concentric spheres with increasing (t > t

0

) or decreasing (t < t

0

) radii

and parallel hyperplanes.

Finally,if k < 0,the space M is hyperbolic.The solutions are given by

a(t) =

p

jkj coth(

p

jkj(t t

0

);a(t) =

p

jkj tanh(

p

jkj(t t

0

);a(t) =

p

jkj:

These solutions correspond to the ve families of equidistant hypersurfaces in the hy-

perbolic space:Concentric spheres with increasing (t > t

0

) or decreasing (t < t

0

) radii,

hypersurfaces which are parallel to an (n 1)-dimensional hyperbolic subspace,and

expanding (t > t

0

) or contracting (t < t

0

) horospheres.

Fig.5.

11

3.Comparison theory.

We want to derive a comparison theorem for solutions of the Riccati equation

A

0

+A

2

+R

V

= 0 (cf.2.7).Fixing an integral curve of V (which is a geodesic) and

identifying all tangent spaces T

(t)

M by parallel displacement (i.e.via an orthonormal

basis (E

i

(t)) of vector elds along which are parallel,i.e.E

0

i

= 0),we consider A(t)

as a self adjoint endomorphism on a single vector space E = T

(0)

M.More generally,

let E be a nite-dimensional real vector space with euclidean inner product h;i.The

space S(E) of self adjoint endomorphisms inherits the inner product

hA;Bi = trace(A B) (3:1)

for A;B 2 S(E).We get a partial ordering on S(E) by putting A B if hAx;xi

hBx;xi for every x 2 E.

Theorem 3.1 (cf.[14],[9]) Let R

1

;R

2

:IR!S(E) be smooth with R

1

R

2

.For

i 2 f1;2g let A

i

:[t

0

;t

i

)!S(E) be a solution of

A

0i

+A

2i

+R

i

= 0 (3:2)

with maximal t

i

2 (t

0

;1].Assume that A

1

(t

0

) A

2

(t

0

).Then t

1

t

2

and A

1

(t)

A

2

(t) on (t

0

;t

1

).

Proof.Let U = A

2

A

1

;then U(t

0

) 0 and

U

0

= A

02

A

01

= A

21

A

22

+R

1

R

2

:(3:3)

We dene S = R

1

R

2

0 and X =

1

2

(A

1

+A

2

);the equation (3.3) takes the form

U

0

= XU +UX +S:(3:4)

We solve (3.4) by the variation of constant method (see [14],pag.211,Remark 1).Let

t

0

= minft

1

;t

2

g and g:(t

0

;t

0

)!S(E) be a non-singular solution of the homogeneous

equation

g

0

= Xg:(3:5)

Now a solution U of (3.4) is obtained as

U = g V g

T

where V veries

V

0

= g

1

S (g

1

)

T

:(3:6)

>From S 0 we get V

0

0;this,combined with V (0) 0,implies that V 0 and

hence U 0.Thus A

1

A

2

on (t

0

;t

0

).Since A

0i

is bounded from above,a singularity

can only be negative (going to 1).So A

1

A

2

implies t

0

= t

1

t

2

.

12

Remark 3.2 Theorem 3.1 still holds if A

1

;A

2

are singular at t

0

,but U = A

2

A

1

has

a continuous extension to 0 with U(0) 0.See [14] for the proof.A similar argument

also shows that t

1

< t

2

if A

1

(t

0

) < A

2

(t

0

);for a dierent proof of this fact see [11],

Lemma 3.1.

The geometric interpretation of Theorem 3.1 is:principal curvatures (i.e.eigenvalues of

the shape operator) of equidistant hypersurfaces decrease faster on the space of larger

curvature.In particular,this is true for Riemannian spheres,as follows by Remark 3.2).

Now we want to nd a comparison theorem for equation (2.6).For A 2 S(E),

denote by

(A) the lowest eigenvalue and by

+

(A) the highest eigenvalue of A.

Theorem 3.3 Let A

1

;A

2

:(t

0

;t

0

)!S(E) such that

+

(A

1

(t))

(A

2

(t)) everywhere:(3:7)

Let J

1

;J

2

:(t

0

;t

0

)!E be nonzero solutions of J

0

i

= A

i

J

i

.Then kJ

1

k=kJ

2

k is

monotoneously decreasing.

Moreover,if

lim

t&t

0

kJ

1

k

kJ

2

k

(t) = 1;(3:8)

then kJ

1

k kJ

2

k.

Equality holds at some t 2 (t

0

;t

0

) i for i = 1;2 we have J

i

= j v

i

on [t

0

;t] for some

constant vector v

i

2 E with Av

i

= v

i

and j

0

= j,where =

+

(A

1

) =

(A

2

).

Proof.Since kJ

i

k is smooth,we can consider

kJ

i

k

0

kJ

i

k

=

< J

0

i

;J

i

>

kJ

i

k

2

=

< A

i

J

i

;J

i

>

< J

i

;J

i

>

2 [

(A

i

);

+

(A

i

)]

so that

log(kJ

1

k)

0

=

kJ

1

k

0

kJ

1

k

+

(A

1

)

(A

2

)

kJ

2

k

0

kJ

2

k

= log(kJ

2

k)

0

;

hence

log

kJ

1

k

kJ

2

k

0

0

which implies that kJ

1

k=kJ

2

k is monotoneously decreasing.

If kJ

1

k=kJ

2

k has the same value 1 at t

0

and t,then kJ

1

k = kJ

2

k on [t

0

;t] and we recieve

J

0

i

= A

i

J

i

= J

i

from which the conclusion follows.

13

We consider the most important special cases due to Rauch and Berger (called

Rauch I and Rauch II in [5]):

Rauch I

Suppose that J

i

for i = 1;2 are solutions of J

00

i

+R

i

J

i

= 0 with

(R

1

)

+

(R

2

) and

J

i

(0) = 0;kJ

0

1

(0)k = kJ

0

2

(0)k:

Then kJ

1

k kJ

2

k up to the rst zero of J

1

.

Rauch II

Suppose that J

i

for i = 1;2 are solutions of J

00

i

+R

i

J

i

= 0 with

(R

1

)

+

(R

2

) and

J

0

i

(0) = 0;kJ

1

(0)k = kJ

2

(0)k:

Then kJ

1

k kJ

2

k up to the rst zero of J

1

.

In fact we apply the theorems 3.1 and 3.3 where in the rst case,A

i

(t) t

1

I as

t!0 and in the second case,A

i

(0) = 0.

Corollary 3.4 Let M be a complete manifold with K 0,p

0

;p

1

2 M and :[0;1]!

M a shortest geodesic segment connecting p

0

and p

1

.Let X?

0

be a parallel vector

eld along .Put p

s

(t) = exptX(s) for all s 2 [0;1].Then

jp

0

(t);p

1

(t)j jp

0

;p

1

j

with equality for some t > 0 only if p

0

;p

1

;p

1

(t);p

0

(t) bound a at totally geodesic

rectangle.Proof.We have

jp

0

(t);p

1

(t)j

Z

1

0

k

@

@s

p

s

(t)kdt

and J

s

(t) =

@

@s

p

s

(t) is a Jacobi eld along the geodesic

s

(t) = p

s

(t) with J

0

s

(0) = 0.

Thus comparing with the euclidean case we get from Rauch II that kJ

s

(t)k kJ

s

(0)k

which shows the inequality.If we have equality at t

1

> 0,the equality discussion of

Theorem3.3 shows that J

s

is parallel along

s

j[0;t

1

].Moreover,the curves s 7!p

s

(t) are

shortest geodesics of constant length for 0 t t

1

.Thus the surface p:(s;t) 7!p

s

(t)

is a at rectangle in M with

D

@s

@p

@s

=

D

@t

@p

@s

=

D

@t

@p

@t

= 0;

so it is also totally geodesic,i.e.covariant derivatives of vector elds tangent to p remain

tangent to p.

14

4.Average comparison theorems.

Now we consider the trace of the Riccati equation A

0

+A

2

+R

V

= 0 for self adjoint

A.Since trace and derivative commute,we get

trace(A)

0

+trace(A

2

) +Ric(V ) = 0:(4:1)

This is unfortunately not a dierential equation for trace(A),because of the term

trace(A

2

).However,put

a =

trace(A)

n 1

:

(Note that A(V ) = D

V

V = 0,so we consider A as an endomorphism on the (n 1)-

dimensional subspace E = V

?

of the tangent space.) Then

A = aI +A

0

;

with trace(A

0

) = 0,so A

0

and I are perpendicular.Hence,

trace(A

2

) = kAk

2

= a

2

kIk

2

+kA

0

k

2

= (n 1)a

2

+kA

0

k

2

and we get,from the trace equation (4.1):

a

0

+a

2

+r = 0 (4:2)

with

r =

1

n 1

kA

0

k

2

+Ric(V )

1

n 1

Ric(V ):

Geometric meaning:a(t) is the mean curvature of S

t

.

Theorem 4.1 Suppose that A:[t

0

;t

1

)!S(E) (t

1

+1maximal) is a solution of

A

0

+A

2

+R = 0 (4:3)

where R:IR!S(E) is given;suppose that for some constant k 2 IR:

(1) trace(R) (n 1)k

(2) trace(A(t

0

)) (n 1)

a(t

0

)

where

a:[t

0

;t

2

)!IR is a solution of

a

0

+

a

2

+k = 0 (4:4)

15

with t

2

+1 maximal.Let

a =

trace(A)

n 1

:(4:5)

Then t

1

t

2

and a(t)

a(t) for t 2 [t

0

;t

1

).

Proof.Apply theorem 3.1 with (R

1

;A

1

;R

2

;A

2

) replaced with (r;a;k;

a).

Remark 4.2 By Remark 3.2,the theorem remains true if A(t)

1

tt

0

I and

a is the

solution of (4.4) with a pole at t

0

,i.e.

a = s

0

=s,where s is the solution of

s

00

+ks = 0;s(t

0

) = 0;s

0

(t

0

) = 1:

Next,let J

1

;:::;J

n1

be a basis of solutions of J

0

= A J,and put

j = det(J

1

;:::;J

n1

):

Since

(J

1

^:::^ J

n1

)

0

=

n1

X

k=1

J

1

^:::^ A J

k

^:::^ J

n1

;

we get

j

0

= (n 1)a j:(4:6)

Geometrically,equation (4.6) says how the volume element of S

t

,namely det(d

t

) (see

page 9 of chapter 2),changes with t.

Theorem 4.3 Let A:[t

0

;t

1

)!S(V ) be given with

a

a;

where a =

1

n1

trace(A),and let j be as above.Choose

j such that

j

0

= (n 1)

a

j:

Then j=

j is monotonously decreasing.

Proof.Apply theorem 3.3 with (A

1

;J

1

;A

2

;J

2

) replaced with ((n 1)a;j;(n 1)

a;

j).

16

5.Bishop - Gromov inequality

Let M be a complete connected Riemannian manifold.By the theorem of Hopf

and Rinow (cf.[29]),any two points p;q 2 M can be connected by a shortest geodesic

,i.e.L( ) = jp;qj.Let S

p

M = fv 2 T

p

M:kvk = 1g be the unit sphere in T

p

M.For

any v 2 S

p

M,we dene

cut(v) = maxft:

v

j

[0;t]

is shortestg:

This denes a function cut:S

p

M!(0;1],the cut locus distance,which is continuous

(cf [5],p.94).Let

C

p

= ftv:v 2 S

p

M;t cut(v)g:(5:1)

This is a closed subset of T

p

M,and its boundary @C

p

(sometimes also exp

p

(@C

p

) M)

is called the cut locus of the point p.It follows from this denition that

B

r

(p) = exp

p

(B

r

(0)) = exp

p

(B

r

(0)\C

p

) 8r > 0:(5:2)

In fact,if we choose q 2 B

r

(p),there exists a shortest geodesic

v

joining p and q;the

length of

v

should be cut(v),hence v 2 C

p

(theorem of Hopf - Rinow).

Example 5.1 On the unit sphere we have cut(v) = for every v.In fact,in every

direction,the geodesic is a meridian,hence it is shortest up to the opposite ("antipodal")

point.

Example 5.2 On the cylinder S

1

IR,we have cut(v) = = cos where is the angle

between v and the S

1

-direction.

Fig.6.

17

There are two ways how a geodesic =

v

:[0;1)!M (where v 2 S

p

M) can

cease to be shortest beyond the parameter t

0

= cut(v) (cf.[5],p.93):Either there exists

a nonzero Jacobi eld J along which vanishes at 0 and t

0

- in this case, (t

0

) is called

a conjugate point of p (cf.Example 5.1),or there exists a second geodesic 6= of

the same length which also connects p and (t

0

) (cf.Example 5.2).Hence q = (t

0

)

is in the cut locus of p = (0) i p is in the cut locus of q.Moreover,there are no

conjugate points on j[0;cut(v)).The conjugate points in turn are the singular values

of the exponential map exp

p

;more precisely,we have:

Lemma 5.3 Let J(t) be the Jacobi eld along

v

dened by J(0) = 0,J

0

(0) = w.Then

we have

d(exp

p

)

tv

:tw = J(t):

In particular,d(exp

p

)

tv

is singular if and only if exp

p

(tv) is a conjugate point of p.

Proof.Let w 2 T

v

T

p

M T

p

M.Then we have

d(exp

p

)

v

:w =

d

ds

s=0

exp

p

(v +sw) =

d

ds

s=0

v+sw

(1):(5:3)

If we let

J(t) =

@

@s

s=0

v+sw

(t);(5:4)

then J is the Jacobi eld along

v

with initial conditions J(0) = 0 and

J

0

(0) =

D

@t

0

@

@s

0

v+sw

(t)

=

D

@s

0

@

@t

0

v+sw

(t)

=

D

@s

0

(v +sw) = w

:

Therefore we get

d(exp

p

)

v

w = J(1);(5:5)

and generally

d(exp

p

)

tv

tw = J(t):(5:6)

18

Remark 5.4 Consequently,on the interior of C

p

,the exponential map exp

p

is injective

and regular,hence a dieomorphism.Note that Int(C

p

) is star-shape,thus it is con-

tractive;hence also its image is contractive.But by Hopf-Rinow,the whole manifold

M is the image of exp

p

:C

p

!M,so the topology of M is given by the image of the

boundary @C

p

.

After these preparations,we come to the main theorem of this section.

Theorem 5.5 Let us consider a manifold M

n

with Ricci curvature satisfying

Ric

n 1

k:

Let

M be the complete simply connected n-manifold with constant curvature k (standard

space of constant curvature k) and

B

r

M the ball of radius r in

M.Then,for all

p 2 M,we have that

VolB

r

(p)

Vol

B

r

&

r

(5:7)

i.e.this quotient is monotonely decresing with r.Moreover,for r!0,the quotient

goes to one.

Corollary 5.6 For any two positive real numbers R > r we have

VolB

R

(p)

VolB

r

(p)

Vol

B

R

Vol

B

r

:(5:8)

Remark 5.7 Corollary 5.6 gives an upper bound for the growth of the metric balls in

M.Moreover,if equality holds for some r < R,then B

R

(p) is isometric to

B

r

(this can

be seen from the proof).

Proof of the theorem.By (5.3) we have

VolB

r

(p) =

Z

B

r

(0)\C

p

det

d(exp

p

)

u

du:(5:9)

Passing to polar coordinates and denoting r(v) = minfr;cut(v)g,we get

VolB

r

(p) =

Z

S

Z

r(v)

0

det

d(exp

p

)

tv

t

n1

dt dv (5:10)

19

where S:= S

1

(0) T

p

M.If we consider a basis e

1

;:::;e

n1

of v

?

T

p

M,then by

Lemma 5.3,

d(exp

p

)

tv

e

i

=

1

t

d(exp

p

)

tv

te

i

=

1

t

J

i

(t);

where J

i

is the Jacobi eld along

v

with J

i

(0) = 0 and J

0

i

(0) = e

i

.Hence

det

d(exp

p

)

tv

=

1

t

n1

det (J

1

(t);:::;J

n1

(t));(5:11)

and equation (5.10) becomes

VolB

r

(p) =

Z

S

Z

r(v)

0

j

v

(t)dt dv (5:12)

where

j

v

(t) = det (J

1

(t);:::;J

n1

(t)):(5:13)

If we put j

v

(t) = 0 for t > cut(v),then by the comparison theorem 4.3 we get

(j

v

=

j) &

on [0;r] and hence

q:=

1

Vol(S)

Z

S

(j

v

=

j)dv

is still monotone.Moreover,

Vol

B

r

=

Z

S

Z

r

0

j(t)dtdv = Vol(S)

Z

r

0

j(t)dt:(5:14)

Therefore we have that

VolB

r

(p)

Vol

B

r

=

R

r

0

q(t)

j(t)dt

R

r

0

j(t)dt

(5:15)

is a monotone decreasing function in r,because the mean of a monotone function on

growing intervals is still monotone.

If r!0,both volumes approximate the euclidean ball volume,hence the quotient

goes to one.

20

6.Toponogov's Triangle Comparison Theorem

Let us x o 2 M and let = jo;j.We already know that near o,precisely in

exp

o

(Int(C

o

)nf0g), is a C

1

function and

(exp

o

(v)) = kvk:(6:1)

Let us consider the unit radial eld V = r.Then S

r

= @B

r

(o) is a family of equidistant

hypersurfaces,as in chapter 2.

Suppose that the sectional curvature K of M is k.If

~

M is the standard space of

sectional curvature k,then,by the comparison theorem 3.1,we get

A

~

A =

s

0

s

I;(6:2)

where s is a solution of s

00

+ ks = 0 with initial dates s(0) = 0;s

0

(0) = 1,and A =

DV = Dr is the Hessian of .(Recall from Example 2.4 that a = s

0

=s is the (unique)

solution of the equation a

0

+a

2

+k = 0 with a pole at t = 0.)

Therefore,

Drj

V

?

s

0

s

I;(6:3)

while

Drj

IRV

= 0;(6:4)

because grows linearly along the integral curves of V.Analogous relations hold for ~:

Dr~j

V

?

=

s

0

s

I (6:5)

Dr~j

IRV

= 0 (6:6)

Now we want to nd a unique estimate for the whole Hessian.To get this we modify

(and analogously ~) suitably:Consider = f ,where f:IR!IR is a function yet to

be determined.Then

Dr(f ) = D((f

0

)r) = f

00

()d r +f

0

()Dr:(6:7)

On V

?

we have f

00

()d r = 0 and Dr (s

0

=s)I;while f

00

()d r = f

00

()I and

Dr = 0 on IR V.If we choose f as a principal function of s,i.e.f

0

= s,then we get

f

00

() = kf() +C;

hence (6.3) and (6.4) give

Dr kI +C (6:8)

21

where C is some xed constant.Analogously,for ~ = f ~,we get from (6.5) and (6.6)

Dr~ = k~I +C:(6:9)

Theorem 6.1 (Toponogov's triangle comparison theorem) [18],[5],[24]

Let M be a complete Riemannian manifold with sectional curvature K k.Let

~

M be

the standard space of constant curvature k.Let p

0

;p

1

;o 2 M,and choose correspond-

ing points ~p

0

;~p

1

;~o 2

~

M.Let be a geodesic from p

0

to p

1

,and

i

a shortest geodesic

from p

i

to o,i = 0;1,all parametrized by arc length,and let ~ ;

~

i

be the corresponding

curves in

~

M,with L( ) = L(~ ) = L and L(

i

) = L(

~

i

).Let us suppose that all the

lengths are smaller than =

p

k,if k > 0.Then we have

jo; (t)j j~o;~ (t)j 8t 2 [0;L]:(6:10)

Fig.7.

Remark 6.2 The hypothesis made on lengths (when k > 0) implies that the geodesics

in

~

M are shortest.

Corollary 6.3 Let

0

=

6

(

0

0

;

0

(0)),

1

=

6

(

0

1

;

0

(L)) and let ~

0

,~

1

the corre-

sponding angles in

~

M.Then

i

~

i

:(6:11)

Proof of the corollary.Let us suppose,by contradiction,that

0

< ~

0

.Suppose rst

that p

0

is not in the cut locus of o.Then exp

o

is invertible near p

0

(cf.Lemma 5.3).

22

Let

t

be the shortest geodesic joining o to (t);the corresponding one

~

t

is a shortest

geodesic in

~

M (for t close to 0),hence

L(

t

) jo; (t)j;

L(

~

t

) = j~o;~ (t)j:

We have

L(

~

t

) = jo;p

0

j +t

d

dt

0

L(

~

t

) +O(t

2

) (6:12)

L(

t

) = jo;p

0

j +t

d

dt

0

L(

t

) +O(t

2

) (6:13)

and,by the rst variation formula for curves (cf.[5],p.5),we get

d

dt

0

L(

t

) = <

0

(0);

0

0

(0) >

d

dt

0

L(

~

t

) = < ~

0

(0);

~

0

0

(0) >:

Since we supposed

0

< ~

0

,for small t we get L(

t

) < L(~

t

),which implies

jo; (t)j L(

t

) < L(~

t

) = j~o;~ (t)j:

Thus,by Toponogov's theorem,we get a contradiction.

If p

0

happens to be a cut locus point of o,we choose o

"

=

0

(") on

0

close to

o.Then certainly p

0

is not in the cut locus of o

"

.Now we put

t

the broken geodesic

j[0;"] [

";t

where

";t

denotes the shortest geodesic from o

"

to (t),and the same

argument holds.

Proof of theorem 6.1

Let us dene = jo;j,~ = j~o;j,and = f ,~ = f ~.Consider the function

= ~ ~ :(6:14)

Hence we have to prove that

0 on [0;L]:(6:15)

23

Fig.8.

We prove (6.15) by contradiction.Suppose that there is t 2 [0;L] such that (t) < 0,

and let m= min

[0;L]

(t) < 0.

We choose k

0

> k suciently close to k and > 0 such that

L <

p

k

0

:(6:16)

It is easy to nd a solution a

0

of the equation a

000

+ k

0

a

0

= 0,with a

0

() = 0 and

a

0

j

[0;L]

m.Then there exists > 0 such that a = a

0

satises the following proper-

ties:

1.a

2.a(t

0

) = (t

0

) for some t

0

2 (0;L).

Case 1: (t

0

) is not a cut locus point of o.Thus is of class C

1

in a neighborhood of

t

0

and

( )

00

=< D

0

r;

0

> k( ) +C;(6:17)

where the inequality follows from (6.8).By eqution (6.9) we get

(~ ~ )

00

=< D

~

0

r~;~

0

>= k(~ ~ ) +C:(6:18)

Hence

00

k:(6:19)

On the other hand a

00

= k

0

a.Moreover,in t

0

we have (t

0

) = a(t

0

) < 0,which implies

( a)

00

(t

0

) (t

0

)(k

0

k) < 0:(6:20)

This is a contradiction because a takes a minimum at t

0

.

24

Case 2: (t

0

) is a cut locus point of o.Let be a shortest geodesic from o to (t

0

).We

choose o

"

on close to o,say jo

"

;oj =".Then we replace by

"

(x):= jx;o

"

j +jo

"

;oj.

By triangle inequality,

"

(x) (x);(6:21)

and equality holds at x = (t

0

).In other words,

"

is an upper support function of at

(t

0

).Since is shortest from o to (t

0

),o

"

is not a cut point of (t

0

),and therefore,

(t

0

) is not a cut point of o

"

(cf.Ch.5).Putting

"

= f

"

,we get the same estimates

as in Case 1 for

"

in place of ,up to a small error which goes to zero as"!0:

(

"

)

00

k(

"

) +C +error:(6:22)

Now

"

is an upper support function of at (t

0

) as f is monotoneously increasing.

Hence

"

a is an upper support function of a at t

0

where

"

=

"

~.Thus it also

takes a minimum at t

0

.But this is a contradiction,because (

"

a)

00

(t

0

) < 0 by (6.20).

Remark 6.4 The above proof is essentially due to Karcher ([24]).Recently,M.Kurzel

([27]) extended this proof to the case where curvature bounds are given which depend

radially on the point (rather than being constant).

25

7.Number of generators and growth of the fundamental group

Let M be a complete Riemannian manifold and

^

M its universal covering.The

fundamental group

1

(M) will be viewed as group of deck transformations acting on

^

M.

In other words,M is the orbit space of a discrete group

=

1

(M) of isometries of

^

M

acting freely on

^

M,i.e.if g 2 with g(p) = p for some p 2 M,then g = 1.

Remark 7.1 The fundamental group of any compact Riemannian manifold M is nitely

generated.Proof.There exists a compact fundamental domain F (see denition below) for the

action of on

^

M;e.g.one may take the so called Dirichlet fundamental domain

F = fx 2

^

M;jx;oj jx;goj 8g 2 g:

We say that g 2 is small if gF\F 6=;,i.e.if the fundamental domains F and

gF are neighbors.If d(F) denotes the diameter of F,i.e.the largest possible distance

within F,then gF B

2d(F)

(o) for all small g,for some xed o 2 F.Since the subsets

g(Int(F)) are all disjoint with equal volume,there can be only nitely many of them

in this ball,hence there exist only nitely many small g.We claim that they form a

set of generators.In fact,let g 2 arbitrary.Choose a geodesic segment from o to

go.Then is covered by nitely many fundamental domains g

0

F;:::;g

N

F where g

0

= 1

and g

N

= g,and g

i1

F,g

i

F are neighbors.Thus g

1

i

g

i1

is small,and hence g is a

composition of small group elements.

Denition 7.2 A closed subset F

^

M is a fundamental domain for a group acting

on

^

M if

(a.) Int(F)\Int(gF) =;8g 6= 1;

(b.) F =

^

M.

For a noncompact manifold M,the fundamental group may have innitely many

generators.The next theorem shows that this does not happen if M has K 0;in fact,

there is an a-priori bound on the number of generators,i.e.the cardinality of a suitably

chosen set of generators:

Theorem 7.3 (Gromov 1978,cf [24])

There exists a number c(n) such that:

(a) the number of generators for

1

(M) is c(n) for any n-dimensional complete

manifold M with curvature K 0.

(b) the number of generators for

1

(M) is c(n)

1+kD

for any n-dimensional compact

manifold M with curvature K k

2

and diameter bounded,diam(M) D.

26

Proof.We prove only part a);the second part is similar,but more technical (see Remark

at the end of the proof).

We dene a"norm"in as follows:

jgj = jp;g(p)j

for some xed p 2

^

M.There exists g

1

2 n f1g with jg

1

j minimal (not necessarily

unique).By induction,we can construct a sequence (g

j

):given g

1

;:::;g

k

,we dene

k

= hg

1

;:::;g

k

i

and choose g

k+1

2 n

k

such that jg

k+1

j has minimum norm in n

k

.To nish the

proof,we only have to show

Claim:

k

= for some k c

0

(n):= 2

p

5

n

.

Proof of the claim:for j > i we have jg

j

j jg

i

j,and moreover

jg

i

(p);g

j

(p)j = jp;g

1

i

g

j

(p)j = jg

1

i

g

j

j jg

j

j

since g

1

i

g

j

2 n

j1

(otherwise g

1

i

g

j

;g

i

2

j1

which would imply that g

j

2

j1

contradicting the choice of g

j

).Now consider the triangle p,p

i

= g

i

(p),p

j

= g

j

(p).

Fig.9.

Let

v

i

be a shortest geodesic from p to p

i

,and

ij

the angle between v

i

and v

j

.The

standard space

~

M of zero curvature is euclidean space IR

n

.Considering the comparison

triangle ~p,~p

i

,~p

j

in

~

M,we have ~

ij

60

for the corresponding angle ~

ij

in ~p.(Note

that ~

ij

is opposite to the largest edge in that triangle.) By Toponogov's theorem then

ij

60

8i 6= j:(7:1)

27

Fig.10.

There are at most 2

p

5

n

vectors that satisfy (7.1).Namely,for any two vectors v

i

,

v

j

of this kind,balls of radii

1

2

are disjoint and their inner half balls are contained in

B

p

5=2

(0),as the gure shows.

Thus,if there are k such vectors,then

Vol B

p

5=2

(0)

k

2

Vol B1

2

hence

k 2

Vol B

p

5

2

Vol B1

2

= 2

p

5

n

:

This nishes the proof of the claim and of the theorem.

Remark 7.4 A much better (but more dicult) estimate was given by U.Abresch (cf.

[1]).Remark 7.5 In Case (b),we use comparison with a hyperbolic triangle (curvature k

2

)

instead of a euclidean one.Since the side lengths are a priori bounded by the diameter

bound D,this is not much dierence.To see that such a bound is necessary,let M be a

compact surface of genus g with constant negative curvature.Then

1

(M

g

) is generated

by 2g elements,hence is not bounded as g!1).Nevertheless the theorem holds,since

either the curvature or the diameter are unbounded as g!1.

Now let us assume that M has Ric 0 rather than K 0.If M is complete and

noncompact,it is an open question whether the fundamental group is nitely gener-

ated.However,for any nitely generated subgroup,the growth of this group is only

polynomial:

28

Denition 7.6 Let be a nitely generated group and G a nite set of generators of

with G = G

1

and 1 2 G.We dene the growth function N(k) (depending on and

G) as follows:

N(k) = ]fg 2 j 9g

1

;:::;g

k

2 G such that g = g

1

::: g

k

g:(7:2)

So N(k) is the number of group elements which can be written as a product of k

elements of G.The dependence of N(k) on G is easy to estimate:If G

0

is another such

generating set,then there are numbers p;q such that any element of G can be expressed

by p elements of G

0

and each element of G

0

by q elements of G.Thus we have

N

0

(k) N(qk);N(k) N

0

(pk):

Theorem 7.7 (Milnor'68,[30])

Let M be a complete manifold with Ric 0 and let

1

(M) any nitely generated

subgroup of the fundamental group.Then the growth of can be estimated by

N(k) < c k

n

:(7:3)

where the constant c depends on

^

M and the chosen set of generators of .

Proof.Let G be a set of generators as above;it has N(1) elements.Fix a point o 2

^

M.

For all g 2 ,let jgj = jo;goj.Put R

0

= maxfjgj;g 2 Gg.Choose some r > 0 small

enough,so that

B

r

(go)\B

r

(o) =;8g 2 n f1g (7:4)

Put R = R

0

+r.Then the family of balls fB

r

(go);g 2 Gg is disjoint and its union is

contained in B

R

(o) so that

Vol

B

R

(o)

N(1) Vol

B

r

(o)

:(7:5)

We can iterate this argument as follows:At the second step,we consider

G

2

:= fg

1

g

2

;g

1

;g

2

2 Gg:

with ](G

2

) = N(2).Then all balls B

r

(go) with g 2 G

2

are disjoint and contained in

B

2R

(o) so that

Vol

B

2R

(o)

N(2) Vol

B

r

(o)

:(7:6)

In general,we obtain that

Vol

B

kR

(o)

N(k) Vol

B

r

(o)

:(7:7)

29

Recall that we have the Bishop - Gromov inequality (cf.Corollary 5.6),

Vol

B

kR

(o)

!

n

k

n

R

n

;

where!

n

denotes the volume of the euclidean unit ball,hence

N(k)

(

!

n

R

n

Vol

B

r

(o)

)

k

n

:(7:8)

30

8.Gromov's estimate of the Betti numbers

Homology is a main tool to measure the complexity of topology.Fix a eld F and

let H

q

(M) denote the q-th singular homology of M with coecients in F.Further,let

H

(M) =

q0

H

q

(M) be the total homology of M.The total Betti number of M is

given by

b(M) = dim

F

H

(M):(8:1)

Theorem 8.1 Gromov,1980 (cf.[15],[1],[28])

There is a constant C(n) such that:

(a.) any complete n-dimensional manifold M with nonnegative curvature K satises

b(M) C(n);(8:2)

(b.) any compact n-dimensional manifold M with curvature K k

2

,and bounded

diameter,diam(M) D,satises

b(M) C(n)

1+kD

:(8:3)

We will give the proof of part (a.),following ideas of Abresch [1] and W.Meyer

[28].(Part (b.) is similar,cf.Remark 7.2.) The proof uses the estimates of Bishop-

Gromov and Toponogov.It can be viewed as an application of some sort of Morse

theory for the distance function (x) = jo;xj where o 2 M is xed.In ordinary Morse

theory,one considers a smooth function f:M!IR with isolated critical points with

nondegenerate Hessian (p critical means that rf(p) = 0),and one observes how the

topology of M

c

= fx 2 M;f(x) < cg is changed as c grows.There are two main facts

in Morse theory (cf [29]):

(1.) If M

b

n M

a

contains no critical points,then M

b

and M

a

are dieomorphic.

(2.) If M

b

nM

a

contains exactly one critical point p,then M

b

is homotopic to M

a

with

a k-cell attached,where k is the index of the Hessian of f at p.

The distance function = jo;j:M!IR is no longer smooth,but we still have the

notions of critical and regular points:

Denition 8.2 A point x 2 M is called a regular point of if there exists v 2 T

x

M

such that

hv;

0

(0)i < 0 (8:4)

for any shortest geodesic from x to o.Any such vextor v is called gradientlike.

31

A point x 2 M is a critical point for if it is non-regular,i.e.if for any v 2 T

x

M there

is a shortest geodesic from x to o such that

hv;

0

(0)i 0:

Remark 8.3 These notions make sense also if the point o is replaced by a closed subset

M.This will be needed in Ch.10.

Fact (1.) is still valid:Since the set of initial vectors of shortest geodesics to o

is closed,the gradientlike vectors form an open subset of TM and moreover a convex

cone at any regular point.Thus we may cover the closure of M

b

n M

a

= B

b

(o) n B

a

(o)

by nitely many open sets with gradientlike vector elds and past them together using

a partition of unity,thus getting a gradientlike vector eld in a neighborhood of the

closure of B

b

(o) n B

a

(o).This has the property that is strictly increasing along its

integral curves.Hence,pushing along the integral curves,we may deform the bigger

ball B

b

(o) into the smaller one B

a

(o).(See Lemma 10.9 for details.) We will use this

in Lemma 8.10 below.

However,Fact (2.) has no meaning and has to be replaced by another idea:Large

balls can be covered by a bounded number of small balls (Bishop-Gromov inequality),

and the jump of the Betti number when passing from a small ball to a large ball can be

controlled using Toponogov's theorem.

First of all,critical points of are not necessarily isolated,but still in some sense,

we have to take only nitely many into account:

Lemma 8.4 Let M be a complete manifold with nonnegative curvature.For any L > 1

there exists a nite number c(n;L) such that there are at most c(n;L) critical points

fq

i

g for satisfying

jo;q

i+1

j Ljo;q

i

j:(8:5)

E.g.for L = 2 we have c(n;2) = 2

p

5

n

.

Proof.Let (q

1

;q

2

;:::) be a maximal sequence satisfying (8.5).For i < j,let be a

shortest geodesic from q

i

to q

j

and put v =

0

(0).Since q

i

is critical,there is a shortest

geodesic c from q

i

to o with the angle =

6

(c

0

(0);v) 90

.Applying Toponogov's

theorem (Corollary 6.3) with the standard space

~

M = IR

n

,we get

~

90

.

Consider rst the limit case

~

= 90

.Let ~ be the angle in ~o.It follows that

cos(~) =

jq

i

;oj

jq

j

;oj

1

L

:

32

Fig.11.

Hence,if

~

90

,

~

o

arccos(

1

L

) =:

0

:

Now we apply Toponogov's theorem backwards for the angle at o,but this time we

consider an arbitrary shortest geodesic from q

i

to q

j

.Then for the angle at o we have

~

0

:

It follows as in Ch.7 that there must be a nite number of such critical points.If L = 2,

we have

0

= 60

and hence c(n;2) = 2

p

5

n

as in the proof of Theorem 7.3.

Corollary 8.5 Given a complete manifold M with nonnegative curvature,all critical

points are contained in a nite ball.

Since we will work with many metric balls in M,we agree on the following conven-

tion:If B = B

r

(p) be a xed ball and > 0,we put B:= B

r

(p).More generally,for

any q 2 M we let B(q):= B

r

(q).

Denition 8.6 Let A C M.We dene the content of A in C as the rank of the

inclusion map on the homology level:

cont(A;C) = rk(i

:H

(A)!H

(C)) (8:6)

Then we dene the content of a metric ball B as:

cont(B) = cont(B;5B):(8:7)

33

Essentially,the content measures the total Betti number of a subset.But it is better

than the Betti number since it has a nice monotonicity property:Note that if A A

0

B

0

B M then

cont(A;B) cont(A

0

;B

0

)

In the spirit of Morse theory,we will observe how the content of balls grows with

the radius.A measure for the number of critical points which are still outside the ball

and which will eventually increase the content is the corank.This denition involves

also critical points of the distance function

p

= jp;j for points p 2 M dierent from o,

called critical for p for short.

Denition 8.7 Let B = B

r

(o) a ball in M,r > 0.For p 2 M,let k(r;p) be the

maximum number of critical points q

1

;:::;q

k

for p such that

1) jp;q

1

j 3Lr

2) jp;q

i+1

j Ljp;q

i

j.

By Lemma (8.3),k(r;p) c(L;n).Then we dene the corank of B as follows:

corank(B) = inffk(r;p) j p 2 5Bg:(8:8)

Not all balls really contribute to the topology,namely the compressible ones:

Denition 8.8 A ball B M is called compressible if there is

~

B =

3

5

B(q) for some

q 2 2B,and a dieomorphism':M!M such that

'j

Mn5B

= id

and

'(B)

~

B

For short:B is compressible into

~

B.Otherwise,the ball B is called incompressible.

Lemma 8.9 Suppose that B is compressible into

~

B,with

~

B =

3

5

B(p),and p 2 2B.

Then

cont(

~

B) cont(B) (8:9)

and

corank(

~

B) corank(B):(8:10)

Proof.Let'be the dieomorphism which compresses B into

~

B.From B '(B) and

'(B)

~

B 5

~

B 5B;(8:11)

34

it follows that cont(

~

B) cont(B).

To show the second relation,put k = corank(B).Let p 5

~

B 5B.Then there

are m k critical points q

1

;:::;q

m

for p with

jp;q

1

j 3Lr 3L

3

5

r;jp;q

i+1

j Ljp;q

i

j:

Hence k(

3

5

r;p) m k which shows corank(

~

B) k.

Lemma 8.10 If B is incompressible,for any p 2 2B there is a critical point for p in

5

~

B n

~

B,where

~

B =

3

5

B(q).

Proof.Otherwise,we could deform B 5

~

B into

~

B while keeping M n 5B xed,so B

would be compressible.

Lemma 8.11 Let B be incompressible,and put

~

B = B(~p) for some

1

5L

and

~p 2

3

2

B.Then

corank(

~

B) corank(B) +1:(8:12)

Proof.Let ~p 2

3

2

B and p 2 5

~

B 2B.By the previous lemma,there is a critical point

q

0

for p in 3B(p) n

3

5

B(p),so we have

3r jp;q

0

j

3

5

r 3Lr:(8:13)

Now let k = corank(B).Then there are m k critical points q

1

;:::;q

m

for p with

jp;q

1

j 3Lr;jp;q

i+1

j Ljp;q

i

j:(8:14)

Thus by (8.13),

jp;q

1

j Ljp;q

0

j;jp;q

0

j 3Lr:(8:15)

Now (8.14) and (8.15) show

k(r;p) m+1 k +1:

Since p 2 5

~

B was arbitrary,we get

corank(

~

B) k +1:

35

Proof of theorem 8.1

Let K = maxfcorank(B) j B ball in Mg.By Lemma 8.4,K c(n;L),and by Corollary

8.5,for a large enough ball B,there exists an isotopy of M that carries M into B,

hence

cont(B) = cont(M;M) = b(M):(8:16)

Gromov's theoremnow follows if we prove that the content of any metric ball is bounded

by a constant.This is done in 5 steps:

Step 1.If corank(B) = K then cont(B) = 1.

Proof:Let p

0

:= p.By Lemma 8.11,B is compressible into B

1

=

3

5

B(p

1

) for some

p

1

2 2B (otherwise,we could increase the corank).Hence,jp

0

;p

1

j < 2r.By Lemma

8.9,corank(B

1

) = K and cont(B

1

) cont(B).Repeating the argument,we compress

B

1

(and hence B) into B

2

=

3

5

B

1

(p

2

) = (

3

5

)

2

B(p

2

) for some p

2

2 2B

1

.Hence jp

1

;p

2

j <

3

5

2r.After N steps,we have compressed B into B

N

= (

3

5

)

N

B(p

N

) with jp

i

;p

i+1

j <

(

3

5

)

i

2r for i = 0;:::;N.So we have for all q 2 B

N

:

jp;qj <

N

X

i=0

(

3

5

)

i

2r < 5r:

Thus B

N

5B.Since the cut locus distance (injectivity radius) is bounded below

on the closure of B (by compactness),we nd some N such that (

3

5

)

N

r is smaller

than this bound which implies that B

N

is dieormorphic to a euclidean ball and hence

contractible.So,cont(B

N

) = 1 since b(B

N

) = 1.On the other hand,

cont(B

N

) cont(B

N1

) ::: cont(B)

which proves cont(B) = 1.

Step 2.Each ball B with cont(B) 2 is either incompressible or it contains an

incompressible ball

~

B with cont(

~

B) cont(B) and corank(

~

B) corank(B).

Proof.Otherwise,we could use the process of Step 1 to show that cont(B) = 1.

Step 3.If B is incompressible,then any ball

~

B = B(p) with

1

5L

and p 2

3

2

B

satises

corank(

~

B) corank(B) +1:(8:17)

Proof.Cf.Lemma 8.11.

Step 4.There is a number (2L)

n(n+1)

10

n(n+1)(n+2)

with the following property:

Suppose that any ball

~

B with corank(

~

B) k has cont(

~

B) a

k

.Then for any ball B

with corank(B) k 1 we have

cont(B) a

k

(8:18)

36

Proof.By Step 2 we may assume that B is incompressible.Let = 1=(5L 10

n+1

).

We cover B with balls B

i

= B(p

i

) with p

i

2 B for i = 1;:::;N such that the balls

1

2

B

i

are disjoint and inside B.Let

1

2

B

1

=

1

2

B(p

1

) the one of smallest volume among

1

2

B

1

;:::;

1

2

B

N

.Since B 2B(p

1

),we have

vol(2B(p

1

)) vol(B)

N

X

i=1

vol(

1

2

B

1

) N vol(

1

2

B

1

) = N vol(

1

2

B(p

1

)):

On the other hand,by Bishop-Gromov (cf.Corollary 5.6),

vol(2B(p

1

))

vol(

1

2

B(p

1

)

)

!

n

(2r)

n

!

n

(

1

2

r)

n

= (

4

)

n

;

thus

N (

4

)

n

= (2L 10

n+2

)

n

:(8:19)

By Step 3,all B

m

i

:= 10

m

B

i

for m= 0;:::;n+1 have corank k and thus cont(B

i

) a

k

.

Since the radii of the B

i

= B

0

i

are very small,we have (by triangle inequality)

B

[

i

B

i

[

i

10

n+1

B

i

5B;

hence

cont(B) cont(

[

i

B

i

;

[

i

10

n+1

B

i

):

We will see below (cf.Appendix) that we may estimate

cont(

[

i

B

i

;

[

i

10

n+1

B

i

)

n+1

X

j=1

X

i

1

<:::<i

j

cont(

j

\

p=1

B

nj+1

i

p

;

j

\

p=1

10B

nj+1

i

p

) (8:20)

where B

m

i

:= 10

m

B

i

.Since

j

\

p=1

B

m

i

p

B

m

i

1

5B

m

i

1

j

\

p=1

10B

m

i

p

;

we get

cont(

j

\

p=1

B

m

i

p

;

j

\

p=1

10B

m

i

p

) cont(B

m

i

1

) a

k

:

37

There are N balls B

m

1

;:::;B

m

N

,so there are at most N

n

intersections B

m

i

1

\:::\B

m

i

j

where j = 1;:::n.Thus

cont(B) N

n+1

a

k

which nishes the proof by (8.19)

Step 5.Let a

k

still denote an upper bound for the content of any ball with corank

k.By Step 1 we may choose a

K

= 1 where K c:= c(L;n) is the biggest possible

corank.Thus by Step 4,a

K1

= ,hence a

K2

=

2

and eventually (by induction),

a

0

=

K

.Hence we get for any ball B

cont(B)

c

which nishes the proof since b(M) = cont(B) for some big ball (cf.(8.16)).

Remark 8.12 The highest known total Betti number for a manifold with K 0 is 2

n

,

the total Betti number of the n-dimensional torus T

n

= S

1

:::S

1

.

38

9.Convexity.

Denition 9.1 Let M be a Riemannian manifold.A continuous function f:M!IR

is called convex if f :I!IR is convex for any geodesic :I!M,and f is called

concave if f is convex.

Theorem 9.2 A continuous function f:M!IR is convex if for any x 2 M and"> 0

there is a smooth lower support function f

x;"

=

~

f:U

x

!IR (i.e.,

~

f f,

~

f(x) = f(x)),

dened in a neighborhood U

x

M,with Dr

~

f(x) > ".

Proof.Suppose there is a geodesic :I!M such that g = f is not convex.Then

there exists a parabola a:IR!IR with:

1) a

00

"

2) a(t

1

) = g(t

1

) and a(t

2

) = g(t

2

)

3) there exists t 2 (t

1

;t

2

) with a(t) < g(t).

Fig.12.

Dene = g a;then takes a maximum at some point t

0

2 (t

1

;t

2

),and we have

(t

0

) > 0.Consider

~

f = f

x;"

with x = (t

0

) and"small enough.Then ~g =

~

f is

a lower support function for g at t

0

,hence (~g a) also takes a maximum at t

0

.But

g

00

(t

0

) > "and a

00

(t

0

) = ",hence

(~g a)

00

(t

0

) > "+"= 0

which is impossible at a maximum point.

Remark 9.3 By the theorem,a C

2

-function f is convex if Drf 0.If f is only

continuous but satises the assumptions of Theorem 9.2,we say Drf 0 in the sense

of support functions.

39

Example 9.4 Fix o 2 M and let (x) = jo;xj.Then Dr = A() > 0 for small > 0 by

example 2.3.However, is not smooth at o,but

2

is smooth (since (exp

o

(v))

2

= hv;vi

if kvk is small,cf.(6.1)),and it satises

Dr(

2

) = 2(d r + Dr) > 0;

hence

2

is convex near o.

Denition 9.5 A closed subset C M is called convex if any geodesic segment in

M with end points on C lies entirely in C.Clearly,if f:M!IR is a convex function,

then the sublevel sets M

a

= fx 2 M j f(x) ag are convex subsets for all a 2 IR.Note

that the notion of convexity depends on the surrounding manifold.For example,in the

cylinder S

1

IR with radius 1,a metric ball of radius r < is not convex,but it is

convex in a slightly larger ball of radius r +"< inside the cylinder.

Denition 9.6 A smooth hypersurface S = @B M is called convex hypersurface if

for the interior unit normal vector eld N,

DN 0:(9:1)

Clearly,if f is smooth and convex,then S = @M

a

is a convex hypersurface unless a is

a minimum of f (note that rf can vanish only at a minimum of a convex function f).

Vice versa,if S = @B is a strictly convex hypersurface,i.e.DN < 0,then the signed

distance function =

S

= j;Sj is concave near S since DN is its Hessian along S

(cf.Ch.2,p.9).Consequently,C = Clos(B) is convex in a neighborhood of C since it

is a sublevel set of the convex function .

Theorem 9.7 Let B be compact and S = @B a convex hypersurface.If K 0 on B,

then = j;Sj is concave on all of B.

Proof.Let x 2 B and a shortest geodesic from x to S.Let p 2 S be its endpoint.

If we were in euclidean space,there would be a support hyperplane for B at p.In M

we have a similar construction:For large R,let

~

S = exp

p

(@B

R

(RN

p

)\U) where U

is a small neighborhood of RN

p

in T

p

M.We will show that

~

S supports B at p,i.e.

~

S\B =;and

~

S\@B = fpg.To this end,we compare the signed distance functions of

S and ~ of

~

S near p.Since N

p

is a common normal vector for S and

~

S,these functions

agree at p up to rst derivatives.The second derivatives (Hessian) of and ~ are given

by the shape operators A = (DN)

p

and

~

A = (D

~

N)

p

of S and

~

S.By convexity,we have

A 0.On the other hand,

~

A =

1

R

I > 0.(In fact,this holds for the euclidean sphere

@B

R

(RN

p

) T

p

M and hence also for

~

S since exp

p

preserves the covariant derivative

40

D at p,cf.Remark 1.5.) Thus A <

~

A.However,the Hessians of and ~ are both

zero in N

p

-direction,so we have no strict inequality.This can be repaired by passing

to functions f and g ~ which have the same level hypersurfaces as and ~,where

we choose the functions f;g:IR!IR monotone with

f(0) = g(0);f

0

(0) = g

0

(0);f

00

(0) < g

00

(0):

Then f and g ~ agree at p up to rst derivatives with Dd(f )

p

< Dd(g ~)

p

,

so we get f g ~ near p (in fact,"<"outside p).In particular,the level set

~

S = fg ~ = 0g lies outside B = ff > 0g.

Fig.13.

Now we consider the signed distance function ~ of

~

S on its full domain,namely

wherever the mapping

S IR!M:(s;t) 7!exp(tN(s))

is invertible,cf.Ch.2,p.8.In particular, is dened and smooth near x:Since is a

shortest curve from p to S,there are no focal points for S on between p and x.Since

~

A > A at p,the focal distance for

~

S along is strictly larger than for S (cf.Remark

3.2),hence x is no focal point for

~

S,and ~ is dened and smooth near x.

Since

~

S lies outside B,any curve from B to

~

S meets S = @B rst,so ~ is an upper

support function for at x.Moreover,by the comparison theorem 3.1 we have

Dr~(x)

1

R+ ~(x)

=:";

so

S

is convex on B by theorem 9.1.

41

Remark 9.8 The proof shows that S = @B need not to be smooth;it is sucient that

there is a unit vector N

p

2 T

p

M such that

~

S

p;R

:= exp

p

(@B

R

(RN

p

)\U) supports B

for any p 2 S,R > 0 and a neighborhood U of RN

p

in T

p

M.For example,this is true

if Clos(B) =:C is convex (with B = Int(C)).To see this,let p 2 @C and consider the

tangent cone

T

p

C = Closfv 2 T

p

M;exp

p

(tv) 2 C for small t > 0g:

By convexity of C,this is a convex cone in T

p

M (since in exponential coordinates,D-

geodesics and @-geodesics,i.e.straight lines,are close to each other near the origin,see

1.5),hence it is contained in some closed half space H

N

= fv 2 T

p

M;hv;Ni 0g.Any

such vector N = N

p

is called an inner normal vector of @C at p,and N

p

is called an

outer normal vector.By strict convexity of

~

S =

~

S

p;R

,the set

~

D = exp

p

(D

R

(RN

p

)\U)

is convex in a neighborhood of p,being a sublevel set of the convex function ~.Thus,

~

D\C is convex near p.But this shows that

~

D\C = fpg,proving that

~

S supports

C at p.Namely,if q = exp

p

v 2

~

D\C,then exp

p

tv 2

~

D\C for all t 2 [0;1],hence

v 2 T

p

C\D

R

(RN

p

) = f0g,so q = p.

42

10.Open manifolds with nonnegative curvature

Theorem 10.1 (Cheeger - Gromoll 1970 [6])

Let M an open (i.e.complete,noncompact) manifold with K 0.There exists a

compact convex submanifold (without boundary) M called soul such that M is

dieomorphic to the normal bundle .

Recall that the normal bundle of a submanifold M is = [

p

p

,where

p

= fv 2 T

p

M j v?T

p

g.The proof needs some preparations.For the moment,

assume only that M is complete and noncompact.

Denition 10.2 A ray is a geodesic dened on [0;1) which is a shortest geodesic

between any two of its points.

Remark 10.3 For any p 2 M there exists a ray starting at p:Consider a sequence

of points q

i

such that jq

i

;pj!1.Consider shortest geodesics

i

from p to q

i

.There

exists a subsequence of the unit tangent vectors converging to some v 2 S

p

M.The

geodesic

v

in the direction of v is a ray.Moreover,if the points q

i

lie on some geodesic

ray ,i.e.q

i

= (t

i

) with t

i

!1,the geodesic ray

v

is called an asymptote of .Note

that asymptotes are not necessarily unique.

Denition 10.4 The Busemann function associated to a ray is dened as

b

(x) = lim

t!1

(jx; (t)j t) (10:1)

In particular,b

( (s)) = lim(j (s); (t)j t) = lim(t s t) = s.Further,since

(t)

= j; (t)j are Lipschitz-continuous functions with Lipschitz constant L = 1,the

same holds for b

(x).Its level sets are called horospheres.

Consider an asymptote

x

of the ray .We dene

b

x;t

(y) = j

x

(t);yj t +b

(x):(10:2)

b

x;t

is smooth in a neighborhood of p since x is not in the cut locus of any point on

x

.

Lemma 10.5 b

x;t

is a support function of b

at x.

Proof.There is a sequence t

i

!1such that

x

= lim

i

where

i

is a shortest geodesic

from x to (t

i

).Then by triangle inequality,we have for any y 2 M:

b

x;t

(y) = j

x

(t);yj t +b

(x)

j

i

(t);yj t +b

(x)

j (t

i

);yj s

i

+b

(x)

43

Fig.14.

where s

i

= jx; (t

i

)j.The sign""means that the error can be made as small as one

wants (as i!1).From

b

(x) jx; (t

i

)j t

i

= s

i

t

i

we get

b

x;t

(y) jy; (t

i

)j t

i

b

(y):

Moreover,b

x;t

(x) = b

(x).So b

x;t

is a support function to b

.

Lemma 10.6 If M is complete,noncompact with K 0,then each Busemann function

b

(x) is concave.

Proof.It follows from the comparison theorem 3.1 with k = 0 that

Drb

x;t

(x)

1

jx;

x

(t)j

!0

as t!1.So by Theorem 9.2,b

is concave.

Corollary 10.7 The superlevel sets

C

t;

= fx 2 M j b

(x) tg (10:3)

are convex sets in M for any t 2 IR.

44

For any point p 2 M we consider

R

p

= frays :[0;1)!M; (0) = pg:(10:4)

Dene the function

b = min

2R

p

b

:(10:5)

(The inmum is a minimum since a limit of rays is a ray.) b is concave since it is the

minimum of concave functions,and therefore,its superlevel sets

C

t

= fx 2 M j b(x) tg:(10:6)

are convex.

Lemma 10.8 C

t

is compact.

Proof.Assume rst t 0.If C

t

were not compact,there whould exist a sequence of

points q

i

!1 in C

t

.Since b(p) = 0,we have p 2 C

t

.For any i,choose a shortest

geodesic segment

i

from p to q

i

;since C

t

is convex,

i

is contained in C

t

.Since C

t

is closed,the limit ray = lim

i

i

again lies in C

t

.But is a ray starting at p,so

2 R

p

.But then cannot be contained in C

t

since b( (s)) b

( (s)) = s can be

made smaller than t,contradiction!

Since C

s

C

t

for s > t,all superlevel sets of b(x) are compact.

Now let t

0

be the maximum value of b (which exists by compactness).Dene

C

1

= C

t

0

.C

1

cannot contain interior points.(In fact,if x 2 C

1

and b(x) = b

(x),

then b

decreases with speed one along an asymptotic ray

x

of .Thus b b

cannot stay maximal near x.) Thus C

1

is a compact convex set of lower dimension.

In general,a compact convex subset C of a Riemannian manifold is a subset of a (non

complete) submanifold M

1

,such that C has nonempty interior relative to M

1

(cf.[5]).

If @C

1

6=;,we consider the distance function

@C

1

on C

1

.Since C

1

is convex,@C

1

is a

convex hypersurface.>From Theorem 9.7,and Remark 9.8 we see that

@C

1

is concave

on C

1

,thus its superlevel sets

C

1

t

= f

@C

1

tg

are convex again.If t

1

is the maximum of

@C

1

,the set

C

2

= C

1

t

1

cannot have interior points in M

1

,thus is again of lower dimension.In this way we

produce a descending chain

C

1

C

2

::: C

k

45

of compact convex subsets with lower and lower dimension.This process ends after

a nite number (say:k) of steps;if we put = C

k

,then is a compact convex set

without boundary:the soul of M.

It remains to show that M is dieomorphic to .In fact,we will show that M is

dieomorphic to a tubular neighborhood of ,say B

r

(),for small r which itself is

dieomorphic to via the exponential map expj.

Lemma 10.9 For small r > 0,there is a dieomorphism':M!B

2r

() with'= id

on B

r

().

Proof.Let

= j;j.We show rst that

has no critical points on M n ,i.e.for

any x 2 M n there is a gradientlike vector v 2 T

x

M for

.

In fact,since x 62 ,there is j 2 f1;:::;k 1g such that x 2 C

j

n C

j+1

.In particular,

there is some t such that x 2 @C

j

t

.Now,since C

j

t

,any geodesic from x to has

initial vector

0

(0) pointing to the interior of C

j

t

.Thus,an outer normal vector for the

convex set C

j

t

is gradientlike.

Now for any x 2 Mn B

r

(),we choose a gradientlike vector v 2 N

x

C

t

and enlarge

it to a gradientlike vector eld V

x

on some neighborhood U

x

.By paracompactness,we

may pass to a locally nite subcovering (U

x

i

)

i=1;2;:::

.Let V

i

= V

x

i

.On B

2r

() n ,we

put V

0

= r

:We have chosen r > 0 so small that

is smooth on B

2r

();this is

possible by compactness of .We may choose a subordinated partition of unity ('

i

)

i0

:

then V =

P

'

i

V

i

is a smooth gradientlike vector eld dened on M n which agrees

to r

on B

r

() n .

Let c

x

be the integral curve of V with c

x

(0) = x.Since the integral curves intersect

@B

r

() transversally (in fact,orthogonally),there is a smooth function

t:M n !IR

with

c

x

(t(x)) 2 @B

r

():(10:7)

We reparametrize the integral curves and put

~c

x

(t) = c

x

(r +t(x) t):

Now let :IR

+

![0;2r) smooth,with (t) = t for t 2 [0;r] and

0

> 0,

and let':M!B

2r

() with

'(x) =

x x 2 B

r

()

~c

x

((r +t(x))) otherwise

Then'maps dieomorphically M onto B

2r

().

46

Fig.15.

Theorem 10.10 (Perelman) Let M be an open manifold of K 0 with soul .For

any p 2 and any two nonzero vectors a 2 T

p

,v 2

p

,there is a totally geodesic

at half plane through p tangent to a and v.

The proof uses the contraction of Sharafutdinov ([32],[35]) which is a continuous

mapping :M! with

j(x);(y)j jx;yj

for all x;y 2 M.In fact, is the limit of an iterated projection onto more and more

smaller and smaller convex sets surrounding ,cf.[35] for details.

Proof of Perelman's theorem:Let

F = exp:!

and put

f(v) = j(v);Fvj

for v 2 ,where :! denotes the projection.Consider

mf(t):= maxff(v);v 2

t

g

where

t

= fv 2 ;kvk = tg.Clearly mf(0) = 0 and mf(t) 0 for all t 0.We

claim that mf(t) is monotonely decreasing so that we get mf 0.Let t > 0 be small

enough such that mf(t) is strictly less than the cut locus distance on .Let v 2

t

so that f(v) = mf(t).Let be the shortest geodesic in from F(v) to p = (v).By

assumption,the prolongation of stays shortest beyond p up to some point q 2 .Let

w 2

q

be the parallel displacement of v from p to q along .

47

Fig.16.

Lemma 10.11 We have f(w) = mf(t) and jFv;Fwj = jp;qj,and p;q;expv;expw span

a totally geodesic at rectangle R.

Proof.By the contraction property of and Rauch II we have

jFv;Fwj j expv;expwj jp;qj;(10:8)

and moreover,by the maximality of f(v) = mf(t),

jFw;qj = f(w) mf(t):(10:9)

On the other hand,Fv,p and q are lined up on a shortest geodesic,so

jFv;qj = mf(t) +jp;qj:(10:10)

So the triangle inequality for Fv,Fw,q yields equality in (10.9) and (10.10) and we

have proved the Lemma,using the equality case of Rauch II (cf.Corollary 3.4).

Now we consider the vector w

0

= (1 ("=t))w with length kw

0

k = kwk "= t ".

Clearly

mf(t ") f(w

0

) = jq;Fw

0

j:

Since R = (p;expv;expw;q) is a at rectangle,

j expv;expw

0

j

2

j expv;expwj

2

+"

2

= jp;qj

2

(1 +(

"

jp;qj

)

2

);

hence

jFv;Fw

0

j j expv;expw

0

j jp;qj +C"

2

(10:11)

48

Fig.17.

with C = (2jp;qj)

1

.Using (10.10),(10.11) and the triangle inequality for Fv,q,Fw

0

we get

mf(t ") jq;Fw

0

j jFv;qj jFv;Fw

0

j mf(t) C"

2

:

This implies that mf is monotonely decreasing and hence zero.Thus f 0 and

(expv) = (v) for any v 2 .(Note that the condition for t = kvk made at the

beginning now is void.) In particular,j expv;expwj = jv;wj for any two normal

vectors v;w on which are parallel along a shortest geodesic on .So the proof is

nished by the equality case of Rauch II (Corollary 3.4).

Remark 10.12 The above theorem also proves a conjecture of Cheeger and Gromoll

saying that the soul must be a point (and hence M must be dieomorphic to IR

n

)

provided that there is a point q 2 M where all sectional curvatures are strictly positive.

In fact,we can connect q to by a shortest geodesic (which is perpendicular to ).If

has positive dimension,i.e.if there is a nonzero tangent vector a of the soul where it

meets ,then by Perelman's theorem,there is a at totally geodesic half plane spanned

by a and .Thus not all curvatures at q = (0) are positive.- We are indepted to V.

Schroeder for communicating Perelman's proof.

Remark 10.13 By a previous result of Strake and Walshap [33],Perelman's theorem

implies that on a small tubular neighborhood of the soul,the projection :B

r

()!

is a Riemannian submersion,i.e.d

x

is an orthogonal projection up to isometric linear

isomorphisms,for any x 2 B

r

().We conjecture that the Sharafutdinov contraction

:M! is smooth and also a Riemannian submersion.

49

11.The sphere theorem.

One of the most celebrated results in Riemannian geometry is the"sphere theorem";

cf.[2],[5],[11],[18],[25],[28].

Theorem 11.1 (Rauch,Berger,Klingenberg)

Let M

n

be a compact,simply connected manifold,with K > 0.Assume that

maxK

minK

< 4:(11:1)

Then M is homeomorphic to S

n

.

Remark 11.2 The estimate 4 is sharp.There are simply connected manifolds with

maxK

minK

= 4

which are not homeomorphic to S

n

,namely the projective spaces over the elds IC and

IH (called ICP

m

and IHP

m

),and the Cayley projective plane.

Remark 11.3 Another type of sphere theorem using a diameter estimate instead of

the upper curvature bound was given by Grove and Shiohama ([21],[28]).

Proof.We may assume (rescaling the metric) that

1

4

< K < 1 (11:2)

so the comparison spaces are the spheres S

2

and S

1

with radii 2 and 1 respectively.We

x p 2 M,and consider geodesics of length on the three manifolds.

Fig.18.

50

By the upper curvature bound,the rst conjugate point comes later than on S

1

,hence

the exponential map exp

p

j

B

(0) is an immersion (local dieomorphism).

By the lower curvature bound,the immersed hypersurface f:= exp

p

j@B

(0) is strictly

concave,i.e.DN < 0 for the exterior unit normal eld N (cf.Theorem 3.1)

Now we need Theorem 11.4 (see below) to nish the proof.Take two copies D

+

,

D

of the unit disk D

n

and identify D

+

with D

(0) =

B

(0) and put S = @B

(0).

By Theorem 11.4,there exists an immersion F:D

!M and a local dieomorphism

':S = @D

+

!@D

such that f = F '.Let

'

= D

+

[

'

D

= (D

+

qD

)= ;(11:3)

where the equivalence relation is given as follows:x 2 @D

+

and y 2 @D

are equivalent

(x y) i y ='(x).

Fig.19.

This is a smooth manifold which is homeomorphic to S

n

(see gure).It is dieomorphic

if and only if'extends to a dieomorphism :D

+

!D

.Now we dene

^

F:

'

!M

by putting

^

FjD

+

= exp

p

;

^

FjD

= F:(11:4)

Then

^

F is a local dieomorphism.It is onto because the image is open and closed and

it is injective,since M is simply connected.Hence

^

F is a dieomorphism.

Remark 11.4 In dimension n = 2 we cannot apply the above proof.However,the

theorem follows from the Gauss-Bonnet formula:Since K > 0,it follows that

0 <

Z

M

K = 2(M);(11:5)

51

thus (M) = (2 2g) > 0,which implies that g = 0 where g is the genus of M.Hence

M is a sphere.

Theorem 11.5 (Gromov,Eschenburg [11])

Let M

n

,n 3,be a complete manifold with K 0.Let S be a compact (n 1)-

dimensional manifold,and f:S!M an"-convex immersion,i.e.we suppose that a

unit normal vector eld N along f satises DN < ".Then f bounds an immersed

convex disk,i.e.there exists an immersion

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