3Geodesics
III Riemannian Geometry
3.6 Applications
This finally puts us in a position to prove something more interesting.
Synge’s theorem
We are first going to prove the following remarkable result relating curvature
and topology:
Theorem
(Synge’s theorem)
.
Every compact orientable Riemannian manifold
(
M, g
) such that
dim M
is even and has
K
(
g
)
>
0 for all planes at
p ∈ M
is
simply connected.
We can see that these conditions are indeed necessary. For example, we can
consider
RP
2
=
S
2
/ ±
1 with the induced metric from
S
2
. Then this is compact
with positive sectional curvature, but it is not orientable. Indeed it is not simply
connected.
Similarly, if we take
RP
3
, then this has odd dimension, and the theorem
breaks.
Finally, we do need strict inequality, e.g. the flat torus is not simply connected.
We first prove a technical lemma.
Lemma.
Let
M
be a compact manifold, and [
α
] a nontrivial homotopy class
of closed curves in M. Then there is a closed minimal geodesic in [α].
Proof.
Since
M
is compact, we can pick some
ε >
0 such that for all
p ∈ M
, the
map exp
p

B(0,p)
is a diffeomorphism.
Let
`
=
inf
γ∈[α]
`
(
γ
). We know that
` >
0, otherwise, there exists a
γ
with
`
(
γ
)
< ε
. So
γ
is contained in some geodesic coordinate neighbourhood, but
then α is contractible. So ` must be positive.
Then we can find a sequence
γ
n
∈
[
α
] with
γ
n
: [0
,
1]
→ M
,
˙γ
constant,
such that
lim
n→∞
`(γ
n
) = `.
Choose
0 = t
0
< t
1
< ··· < t
k
= 1
such that
t
i+1
− t
i
<
ε
2`
.
So it follows that
d(γ
n
(t
i
), γ
n
(t
i+1
)) < ε
for all
n
sufficiently large and all
i
. Then again, we can replace
γ
n

[t
i
,t
i+1
]
by a
radial geodesic without affecting the limit lim `(γ
n
).
Then we exploit the compactness of
M
(and the unit sphere) again, and pass
to a subsequence of
{γ
n
}
so that
γ
n
(
t
i
)
, ˙γ
n
(
t
i
) are all convergent for every fixed
i as n → ∞. Then the curves converges to some
γ
n
→ ˆγ ∈ [α],
given by joining the limits
lim
n→∞
γ
n
(
t
i
). Then we know that the length
converges as well, and so we know
ˆγ
is minimal among curves in [
α
]. So
ˆγ
is
locally minimal, hence a geodesic. So we can take γ = ˆγ, and we are done.
Proof of Synge’s theorem.
Suppose
M
satisfies the hypothesis, but
π
1
(
M
)
6
=
{
1
}
.
So there is a path
α
with [
α
]
6
= 1, i.e. it cannot be contracted to a point. By the
lemma, we pick a representative γ of [α] that is a closed, minimal geodesic.
We now prove the theorem. We may wlog assume
˙γ
= 1, and
t
ranges in
[0, T ]. Consider a vector field X(t) for 0 ≤ t ≤ T along γ(t) such that
∇X
dt
= 0, g(X(0), ˙γ(0)) = 0.
Note that since g is a geodesic, we know
g(X(t), ˙γ(t)) = 0,
for all
t ∈
[0
, T
] as parallel transport preserves the inner product. So
X
(
T
)
⊥
˙γ(T ) = ˙γ(0) since we have a closed curve.
We consider the map
P
that sends
X
(0)
7→ X
(
T
). This is a linear isometry
of (
˙γ
(0))
⊥
with itself that preserves orientation. So we can think of
P
as a map
P ∈ SO(2n − 1),
where
dim M
= 2
n
. It is an easy linear algebra exercise to show that every
element of
SO
(2
n −
1) must have an eigenvector of eigenvalue 1. So we can find
v ∈ T
p
M
such that
v ⊥ ˙γ
(0) and
P
(
v
) =
v
. We take
X
(0) =
v
. Then we have
X(T ) = v.
Consider now a variation
H
(
t, s
) inducing this
X
(
t
). We may assume
˙γ
s

is
constant. Then
d
ds
`(γ
s
)
s=0
= 0
as
γ
is minimal. Moreover, since it is a minimum, the second derivative must be
positive, or at least nonnegative. Is this actually the case?
We look at the second variation formula of length. Using the fact that the
loop is closed, the formula reduces to
d
2
ds
2
`(γ
s
)
s=0
= −
Z
T
0
R(X, ˙γ, X, ˙γ) dt.
But we assumed the sectional curvature is positive. So the second variation is
negative! This is a contradiction.
Conjugate points
Recall that when a geodesic starts moving, for a short period of time, it is
lengthminimizing. However, in general, if we keep on moving for a long time,
then we cease to be minimizing. It is useful to characterize when this happens.
As before, for a vector field J along a curve γ(t), we will write
J
0
=
∇J
dt
.
Definition (Conjugate points). Let γ(t) be a geodesic. Then
p = γ(α), q = γ(β)
are conjugate points if there exists some nontrivial
J
such that
J
(
α
) = 0 =
J
(
β
).
It is easy to see that this does not depend on parametrization of the curve,
because Jacobi fields do not.
Proposition.
(i)
If
γ
(
t
) =
exp
p
(
ta
), and
q
=
exp
p
(
βa
) is conjugate to
p
, then
q
is a singular
value of exp.
(ii) Let J be as in the definition. Then J must be pointwise normal to ˙γ.
Proof.
(i)
We wlog [
α, β
] = [0
,
1]. So
J
(0) = 0 =
J
(1). We
a
=
˙γ
(0) and
w
=
J
0
(0).
Note that
a, w
are both nonzero, as Jacobi fields are determined by initial
conditions. Then q = exp
p
(a).
We have shown earlier that if J(0) = 0, then
J(t) = (d exp
p
)
ta
(tw)
for all 0
≤ t ≤
1. So it follows (d
exp
p
)
a
(
w
) =
J
(1) = 0. So (d
exp
p
)
a
has
nontrivial kernel, and hence isn’t surjective.
(ii) We claim that any Jacobi field J along a geodesic γ satisfies
g(J(t), ˙γ(t)) = g(J
0
(0), ˙γ(0))t + g(J(0), ˙γ(0)).
To prove this, we note that by the definition of geodesic and Jacobi fields,
we have
d
dt
g(J
0
, ˙γ) = g(J
00
, ˙γ(0)) = −g(R( ˙γ, J), ˙γ, ˙γ) = 0
by symmetries of R. So we have
d
dt
g(J, ˙γ) = g(J
0
(t), ˙γ(t)) = g(J
0
(0), ˙γ(0)).
Now integrating gives the desired result.
This result tells us g(J(t), ˙γ(t)) is a linear function of t. But we have
g(J(0), ˙γ(0)) = g(J(1), ˙γ(1)) = 0.
So we know g(J(t), ˙γ(t)) is constantly zero.
From the proof, we see that for any Jacobi field with J(0) = 0, we have
g(J
0
(0), ˙γ(0)) = 0 ⇐⇒ g(J(t), ˙γ(t)) = constant.
This implies that the dimension of the normal Jacobi fields along
γ
satisfying
J(0) = 0 is dim M − 1.
Example.
Consider
M
=
S
2
⊆ R
3
with the round metric, i.e. the “obvious”
metric induced from
R
3
. We claim that
N
= (0
,
0
,
1) and
S
= (0
,
1
,
0) are
conjugate points.
To construct a Jacobi field, instead of trying to mess with the Jacobi equation,
we construct a variation by geodesics. We let
f(t, s) =
cos s sin t
sin s sin t
cos t
.
We see that when
s
= 0, this is the greatcircle in the (
x, z
)plane. Then we have
a Jacobi field
J(t) =
∂f
∂s
s=0
=
0
sin t
0
.
This is then a Jacobi field that vanishes at N and S.
p
When we are at the conjugate point, then there are many adjacent curves
whose length is equal to ours. If we extend our geodesic beyond the conjugate
point, then it is no longer even locally minimal:
p
q
We can push the geodesic slightly over and the length will be shorter. On the
other hand, we proved that up to the conjugate point, the geodesic is always
locally minimal.
In turns out this phenomenon is generic:
Theorem.
Let
γ
: [0
,
1]
→ M
be a geodesic with
γ
(0) =
p
,
γ
(1) =
q
such that
p
is conjugate to some
γ
(
t
0
) for some
t
0
∈
(0
,
1). Then there is a piecewise smooth
variation of f(t, s) with f(t, 0) = γ(t) such that
f(0, s) = p, f(1, s) = q
and `(f( ·, s)) < `(γ) whenever s 6= 0 is small.
The proof is a generalization of the example we had above. We know that up
to the conjugate point, we have a Jacobi filed that allows us to vary the geodesic
without increasing the length. We can then give it a slight “kick” and then the
length will decrease.
Proof.
By the hypothesis, there is a
J
(
t
) defined on
t ∈
[0
,
1] and
t
0
∈
(0
,
1) such
that
J(t) ⊥ ˙γ(t)
for all t, and J(0) = J(t
0
) = 0 and J 6≡ 0. Then J
0
(t
0
) 6= 0.
We define a parallel vector field
Z
1
along
γ
by
Z
1
(
t
0
) =
−J
0
(
t
0
). We pick
θ ∈ C
∞
[0, 1] such that θ(0) = θ(1) = 0 and θ(t
0
) = 1.
Finally, we define
Z = θZ
1
,
and for α ∈ R, we define
Y
α
(t) =
(
J(t) + αZ(t) 0 ≤ t ≤ t
0
αZ(t) t
0
≤ t ≤ 1
.
We notice that this is not smooth at
t
0
, but is just continuous. We will postpone
the choice of α to a later time.
We know
Y
α
(
t
) arises from a piecewise
C
∞
variation of
γ
, say
H
α
(
t, s
). The
technical claim is that the second variation of length corresponding to
Y
α
(
t
) is
negative for some α.
We denote by
I
(
X, Y
)
T
the symmetric bilinear form that gives rise to the
second variation of length with fixed end points. If we make the additional
assumption that
X, Y
are normal along
γ
, then the formula simplifies, and
reduces to
I(X, Y )
T
=
Z
T
0
(g(X
0
, Y
0
) − R(X, ˙γ, Y, ˙γ)) dt.
Then for H
α
(t, s), we have
d
2
ds
2
`(γ
s
)
s=0
= I
1
+ I
2
+ I
3
I
1
= I(J, J)
t
0
I
2
= 2αI(J, Z)
t
0
I
3
= α
2
I(Z, Z)
1
.
We look at each term separately.
We first claim that I
1
= 0. We note that
d
dt
g(J, J
0
) = g(J
0
, J
0
) + g(J, J
00
),
and
g
(
J, J
00
) added to the curvature vanishes by the Jacobi equation. Then
by integrating by parts and applying the boundary condition, we see that
I
1
vanishes.
Also, by integrating by parts, we find
I
2
= 2αg(Z, J
0
)
t
0
0
.
Whence
d
2
ds
2
`(γ
s
)
s=0
= −2αJ
0
(t
0
)
2
+ α
2
I(Z, Z)
1
.
Now if
α >
0 is very very small, then the linear term dominates, and this is
negative. Since the first variation vanishes (
γ
is a geodesic), we know this is a
local maximum of length.
Note that we made a compromise in the theorem by doing a piecewise
C
∞
variation instead of a smooth one, but of course, we can fix this by making a
smooth approximation.
Bonnet–Myers diameter theorem
We are going to see yet another application of our previous hard work, which
may also be seen as an interplay between curvature topology. In case it isn’t
clear, all our manifolds are connected.
Definition (Diameter). The diameter of a Riemannian manifold (M, g) is
diam(M, g) = sup
p,q∈M
d(p, q).
Of course, this definition is valid for any metric space.
Example. Consider the sphere
S
n−1
(r) = {x ∈ R
n
: x = r},
with the induced “round” metric. Then
diam(S
n−1
(r)) = πr.
It is an exercise to check that
K ≡
1
r
2
.
We will also need the following notation:
Notation.
Let
h,
ˆ
h
be two symmetric bilinear forms on a real vector space. We
say h ≥
ˆ
h if h −
ˆ
h is nonnegative definite.
If
h,
ˆ
h ∈
Γ(
S
2
T
∗
M
) are fields of symmetric bilinear forms, we write
h ≥
ˆ
h
if
h
p
≥
ˆ
h
p
for all p ∈ M.
The following will also be useful:
Definition
(Riemannian covering map)
.
Let (
M, g
) and (
˜
M, ˜g
) be two Rieman
nian manifolds, and
f
:
˜
M → M
be a smooth covering map. We say
f
is a
Riemannian covering map if it is a local isometry. Alternatively,
f
∗
g
=
˜g
. We
say
˜
M is a Riemannian cover of M .
Recall that if
f
is in fact a universal cover, i.e.
˜
M
is simply connected, then
we can (noncanonically) identify π
1
(M) with f
−1
(p) for any point p ∈ M.
Definition
(Bonnet–Myers diameter theorem)
.
Let (
M, g
) be a complete
n

dimensional manifold with
Ric(g) ≥
n − 1
r
2
g,
where r > 0 is some positive number. Then
diam(M, g) ≤ diam S
n
(r) = πr.
In particular, M is compact and π
1
(M) is finite.
Proof.
Consider any
L < diam
(
M, g
). Then by definition (and Hopf–Rinow),
we can find
p, q ∈ M
such that
d
(
p, q
) =
L
, and a minimal geodesic
γ ∈
Ω(
p, q
)
with `(γ) = d(p, q). We parametrize γ : [0, L] → M so that ˙γ = 1.
Now consider any vector field
Y
along
γ
such that
Y
(
p
) = 0 =
Y
(
q
). Since
Γ is a minimal geodesic, it is a critical point for
`
, and the second variation
I
(
Y, Y
)
[0,L]
is nonnegative (recall that the second variation has fixed end points).
We extend
˙γ
(0) to an orthonormal basis of
T
p
M
, say
˙γ
(0) =
e
1
, e
2
, ··· , e
n
.
We further let X
i
be the unique vector field such that
X
0
i
= 0, X
i
(0) = e
i
.
In particular, X
1
(t) = ˙γ(t).
For i = 2, ··· , n, we put
Y
i
(t) = sin
πt
L
X
i
(t).
Then after integrating by parts, we find that we have
I(Y
i
, Y
i
)
[0,L]
= −
Z
L
0
g(Y
00
i
+ R( ˙γ, Y
i
)Y
i
, ˙γ) dt
Using the fact that X
i
is parallel, this can be written as
=
Z
L
0
sin
2
πt
L
π
2
L
2
− R( ˙γ, X
i
, ˙γ, X
i
)
dt,
and since this is length minimizing, we know this is ≥ 0.
We note that we have R( ˙γ, X
1
, ˙γ, X
1
) = 0. So we have
n
X
i=2
R( ˙γ, X
i
, ˙γ, X
i
) = Ric( ˙γ, ˙γ).
So we know
n
X
i=2
I(Y
i
, Y
i
) =
Z
L
0
sin
2
πt
L
(n − 1)
π
2
L
− Ric( ˙γ, ˙γ)
dt ≥ 0.
We also know that
Ric( ˙γ, ˙γ) ≥
n − 1
r
2
by hypothesis. So this implies that
π
2
L
2
≥
1
r
2
.
This tells us that
L ≤ πr.
Since L is any number less that diam(M, g), it follows that
diam(M, g) ≤ πr.
Since
M
is known to be complete, by HopfRinow theorem, any closed bounded
subset is compact. But M itself is closed and bounded! So M is compact.
To understand the fundamental, group, we simply have to consider a universal
Riemannian cover
f
:
˜
M → M
. We know such a topological universal covering
space must exist by general existence theorems. We can then pull back the
differential structure and metric along
f
, since
f
is a local homeomorphism.
So this gives a universal Riemannian cover of
M
. But they hypothesis of the
theorem is local, so it is also satisfied for
˜
M
. So it is also compact. Since
f
−1
(
p
)
is a closed discrete subset of a compact space, it is finite, and we are done.
It is an easy exercise to show that the hypothesis on the Ricci curvature
cannot be weakened to just saying that the Ricci curvature is positive definite.
Hadamard–Cartan theorem
To prove the next result, we need to talk a bit more about coverings.
Proposition.
Let (
M, g
) and (
N, h
) be Riemannian manifolds, and suppose
M
is complete. Suppose there is a smooth surjection
f
:
M → N
that is a local
diffeomorphism. Moreover, suppose that for any
p ∈ M
and
v ∈ T
p
M
, we have
df
p
(v)
h
≥ v. Then f is a covering map.
Proof.
By general topology, it suffices to prove that for any smooth curve
γ
: [0
,
1]
→ N
, and any
q ∈ M
such that
f
(
q
) =
γ
(0), there exists a lift of
γ
starting from from q.
M
[0, 1] N
f
γ
˜γ
From the hypothesis, we know that
˜γ
exists on [0
, ε
0
] for some “small”
ε
0
>
0.
We let
I = {0 ≤ ε ≤: ˜γ exists on [0, ε]}.
We immediately see this is nonempty, since it contains
ε
0
. Moreover, it is not
difficult to see that
I
is open in [0
,
1], because
f
is a local diffeomorphism. So it
suffices to show that I is closed.
We let {t
n
}
∞
n=1
⊆ I be such that t
n+1
> t
n
for all n, and
lim
n→∞
t
n
= ε
1
.
Using HopfRinow, either
{˜γ
(
t
n
)
}
is contained in some compact
K
, or it is
unbounded. We claim that unboundedness is impossible. We have
`(γ) ≥ `(γ
[0,t
n
]
) =
Z
t
n
0
˙γ dt
=
Z
t
n
0
df
˜γ(t)
˙
˜γ(t) dt
≥
Z
t
n
0

˙
˜γ dt
= `(˜γ
[0,t
n
]
)
≥ d(˜γ(0), ˜γ(t
n
)).
So we know this is bounded. So by compactness, we can find some
x
such that
˜γ
(
t
n
`
)
→ x
as
` → ∞
. There exists an open
x ∈ V ⊆ M
such that
f
V
is a
diffeomorphism.
Since there are extensions of
˜γ
to each
t
n
, eventually we get an extension to
within
V
, and then we can just lift directly, and extend it to
ε
1
. So
ε
1
∈ I
. So
we are done.
Corollary.
Let
f
:
M → N
be a local isometry onto
N
, and
M
be complete.
Then f is a covering map.
Note that since
N
is (assumed to be) connected, we know
f
is necessarily
surjective. To see this, note that the completeness of
M
implies completeness of
f
(
M
), hence
f
(
M
) is closed in
N
, and since it is a local isometry, we know
f
is
in particular open. So the image is open and closed, hence f(M) = N.
For a change, our next result will assume a negative curvature, instead of a
positive one!
Theorem
(Hadamard–Cartan theorem)
.
Let (
M
n
, g
) be a complete Riemannian
manifold such that the sectional curvature is always nonpositive. Then for every
point
p ∈ M
, the map
exp
p
:
T
p
M → M
is a covering map. In particular, if
π
1
(M) = 0, then M is diffeomorphic to R
n
.
We will need one more technical lemma.
Lemma.
Let
γ
(
t
) be a geodesic on (
M, g
) such that
K ≤
0 along
γ
. Then
γ
has no conjugate points.
Proof.
We write
γ
(0) =
p
. Let
I
(
t
) be a Jacobi field along
γ
, and
J
(0) = 0. We
claim that if J is not identically zero, then J does not vanish everywhere else.
We consider the function
f(t) = g(J(t), J(t)) = J(t)
2
.
Then f(0) = f
0
(0) = 0. Consider
1
2
f
00
(t) = g(J
00
(t), J(t)) + g(J
0
(t), J
0
(t)) = g(J
0
, J
0
) − R( ˙γ, J, ˙γ, J) ≥ 0.
So f is a convex function, and so we are done.
We can now prove the theorem.
Proof of theorem.
By the lemma, we know there are no conjugate points. So
we know
exp
p
is regular everywhere, hence a local diffeomorphism by inverse
function theorem. We can use this fact to pull back a metric from
M
to
T
p
M
such that
exp
p
is a local isometry. Since this is a local isometry, we know
geodesics are preserved. So geodesics originating from the origin in
T
p
M
are
straight lines, and the speed of the geodesics under the two metrics are the same.
So we know
T
p
M
is complete under this metric. Also, by Hopf–Rinow,
exp
p
is
surjective. So we are done.