6The Cheeger–Gromoll splitting theorem

III Riemannian Geometry

6 The Cheeger–Gromoll splitting theorem

We will talk about the Cheeger–Gromoll splitting theorem. This is a hard

theorem, so we will not prove it. However, we will state it, and discuss a bit

about it. To state the theorem, we need some preparation.

Definition

(Ray)

.

Let (

M, g

) be a Riemannian manifold. A ray is a map

r

(

t

) : [0

, ∞

)

→ M

if

r

is a geodesic, and minimizes the distance between any

two points on the curve.

Definition

(Line)

.

A line is a map

`

(

t

) :

R → M

such that

`

(

t

) is a geodesic,

and minimizes the distance between any two points on the curve.

We have seen from the first example sheet that if

M

is a complete unbounded

manifold, then

M

has a ray from each point, i.e. for all

x ∈ M

, there exists a

ray r such that r(0) = x.

Definition

(Connected at infinity)

.

A complete manifold is said to be connected

at infinity if for all compact set

K ⊆ M

, there exists a compact

C ⊇ K

such

that for every two points

p, q ∈ M \ C

, there exists a path

γ ∈

Ω(

p, q

) such that

γ(t) ∈ M \ K for all t.

We say M is disconnected at infinity if it is not connected at infinity.

Note that if

M

is disconnected at infinity, then it must be unbounded, and

in particular non-compact.

Lemma. If M is disconnected at infinity, then M contains a line.

Proof.

Note that

M

is unbounded. Since

M

is disconnected at infinity, we can

find a compact subset

K ⊆ M

and sequences

p

m

, q

m

→ ∞

as

m → ∞

(to make

this precise, we can pick some fixed point

x

, and then require

d

(

x, p

m

)

, d

(

x, q

m

)

→

∞) such that every γ

m

∈ Ω(p

m

, q

m

) passes through K.

In particular, we pick

γ

m

to be a minimal geodesic from

p

m

to

q

m

parametrized

by arc-length. Then

γ

m

passes through

K

. By reparametrization, we may assume

γ

m

(0) ∈ K.

Since

K

is compact, we can pass to a subsequence, and wlog

γ

m

(0)

→ x ∈ K

and ˙γ

m

(0) → a ∈ T

x

M (suitably interpreted).

Then we claim the geodesic

γ

x,a

(

t

) is the desired line. To see this, since

solutions to ODE’s depend smoothly on initial conditions, we can write the line

as

`(t) = lim

m→∞

γ

m

(t).

Then we know

d(`(s), `(t)) = lim

m→∞

d(γ

m

(s), γ

m

(t)) = |s − t|.

So we are done.

Let’s look at some examples.

Example. The elliptic paraboloid

{z = x

2

+ y

2

} ⊆ R

3

with the induced metric does not contain a line. To prove this, we can show that

any geodesic that is not a meridian must intersect itself.

Example.

Any complete metric

g

on

S

n−1

× R

contains a line since it is

disconnected at ∞.

Theorem

(Cheeger–Gromoll line-splitting theorem (1971))

.

If (

M, g

) is a com-

plete connected Riemannian manifold containing a line, and has

Ric

(

g

)

≥

0 at

each point, then

M

is isometric to a Riemannian product (

N × R, g

0

+ d

t

2

) for

some metric g

0

on N .

We will not prove this, but just see what the theorem can be good for.

First of all, we can slightly improve the statement. After applying the

theorem, we can check again if

N

contains a line or not. We can keep on

proceeding and splitting lines off. Then we get

Corollary.

Let (

M, g

) be a complete connected Riemannian manifold with

Ric

(

g

)

≥

0. Then it is isometric to

X × R

q

for some

q ∈ N

and Riemannian

manifold X, where X is complete and does not contain any lines.

Note that if

X

is zero-dimensional, since we assume all our manifolds are

connected, then this implies

M

is flat. If

dim X

= 1, then

X

∼

=

S

1

(it can’t be a

line, because a line contains a line). So again M is flat.

Now suppose that in fact

Ric

(

g

) = 0. Then it is not difficult to see from the

definition of the Ricci curvature that

Ric

(

X

) = 0 as well. If we know

dim X ≤

3,

then M has to be flat, since in dimensions ≤ 3, the Ricci curvature determines

the full curvature tensor.

We can say a bit more if we assume more about the manifold. Recall (from

example sheets) that a manifold is homogeneous if the group of isometries

acts transitively. In other words, for any

p, q ∈ M

, there exists an isometry

φ

:

M → M

such that

φ

(

p

) =

q

. This in particular implies the metric is complete.

It is not difficult to see that if

M

is homogeneous, then so is

X

. In this case,

X

must be compact. Suppose not. Then

X

is unbounded. We will obtain a line

on X.

By assumption, for all

n

= 1

,

2

, ···

, we can find

p

n

, q

n

with

d

(

p

n

, q

n

)

≥

2

n

.

Since

X

is complete, we can find a minimal geodesic

γ

n

connecting these two

points, parametrized by arc length. By homogeneity, we may assume that the

midpoint

γ

n

(0) is at a fixed point

x

0

. By passing to a subsequence, we wlog

˙γ

n

(0) converges to some

a ∈ T

x

0

(

X

. Then we use

a

as an initial condition for

our geodesic, and this will be a line.

A similar argument gives

Lemma.

Let (

M, g

) be a compact Riemannian manifold, and suppose its uni-

versal Riemannian cover (

˜

M, ˜g) is non-compact. Then (

˜

M, ˜g) contains a line.

Proof.

We first find a compact

K ⊆

˜

M

such that

π

(

K

) =

M

. Since

˜

M

must be

complete, it is unbounded. Choose

p

n

, q

n

, γ

n

like before. Then we can apply deck

transformations so that the midpoint lies inside

K

, and then use compactness of

K to find a subsequence so that the midpoint converges.

We do more applications.

Corollary.

Let (

M, g

) be a compact, connected manifold with

Ric

(

g

)

≥

0. Then

–

The universal Riemannian cover is isometric to the Riemannian product

X ×R

N

, with X compact, π

1

(X) = 1 and Ric(g

X

) ≥ 0.

– If there is some p ∈ M such that Ric(g)

p

> 0, then π

1

(M) is finite.

–

Denote by

I

(

˜

M

) the group of isometries

˜

M →

˜

M

. Then

I

(

˜

M

) =

I

(

X

)

×

E(R

q

), where E(R

q

) is the group of rigid Euclidean motions,

y 7→ Ay + b

where b ∈ R

n

and A ∈ O(q).

– If

˜

M is homogeneous, then so is X.

Proof.

– This is direct from Cheeger–Gromoll and the previous lemma.

–

If there is a point with strictly positive Ricci curvature, then the same is

true for the universal cover. So we cannot have any non-trivial splitting.

So by the previous part,

˜

M

must be compact. By standard topology,

|π

1

(M)| = |π

−1

({p})|.

–

We use the fact that

E

(

R

q

) =

I

(

R

q

). Pick a

g ∈ I

(

˜

M

). Then we know

g

takes lines to lines. Now use that all lines in

˜

M ×R

q

are of the form

p ×R

with p ∈ X and R ⊆ R

q

an affine line. Then

g(p × R) = p

0

× R,

for some

p

0

and possibly for some other copy of

R

. By taking unions, we

deduce that g(p × R

q

) = p

0

× R

q

. We write h(p) = p

0

. Then h ∈ I(X).

Now for any

X × a

with

a ∈ R

q

, we have

X × a ⊥ p × R

q

for all

p ∈ X

.

So we must have

g(X × a) = X × b

for some b ∈ R

q

. We write e(a) = b. Then

g(p, a) = (h(p), e(a)).

Since the metric of

X

and

R

q

are decoupled, it follows that

h

and

e

must

separately be isometries.

We can look at more examples.

Proposition.

Consider

S

n

× R

for

n

= 2 or 3. Then this does not admit any

Ricci-flat metric.

Proof.

Note that

S

n

× R

is disconnected at

∞

. So any metric contains a line.

Then by Cheeger–Gromoll,

R

splits as a Riemannian factor. So we obtain

Ric

= 0 on the

S

n

factor. Since we are in

n

= 2

,

3, we know

S

n

is flat, as the

Ricci curvature determines the full curvature. So

S

n

is the quotient of

R

n

by a

discrete group, and in particular π

1

(S

n

) 6= 1. This is a contradiction.

Let

G

be a Lie group with a bi-invariant metric

g

. Suppose the center

Z

(

G

)

is finite. Then the center of

g

is trivial (since it is the Lie algebra of

G/Z

(

G

),

which has trivial center). From sheet 2, we find that

Ric

(

g

)

>

0 implies

π

1

(

G

) is

finite. The converse is also true, but is harder. This is done on Q11 of sheet 3 —

if π

1

(G) is finite, then Z(G) is finite.