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
n1
× 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.