3Geodesics
III Riemannian Geometry
3.3 Further properties of geodesics
We can now use Jacobi fields to prove interesting things. We now revisit the
Gauss lemma, and deduce a stronger version.
Lemma (Gauss’ lemma). Let a, w ∈ T
p
M, and
γ = γ
p
(t, a) = exp
p
(ta)
a geodesic. Then
g
γ(t)
((d exp
p
)
ta
a, (d exp
p
)
ta
w) = g
γ(0)
(a, w).
In particular,
γ
is orthogonal to
exp
p
{v ∈ T
p
M
:
v
=
r}
. Note that the latter
need not be a submanifold.
This is an improvement of the previous version, which required us to live in
the geodesic local coordinates.
Proof. We fix any r > 0, and consider the Jacobi field J satisfying
J(0) = 0, J
0
(0) =
w
r
.
Then by the corollary, we know the Jacobi field is
J(t) = (d exp
p
)
ta
tw
r
.
We may write
w
r
= λa + u,
with
a ⊥ u
. Then since Jacobi fields depend linearly on initial conditions, we
write
J(t) = λt ˙γ(t) + J
n
(t)
for a Jacobi field J
n
a normal vector field along γ. So we have
g(J(r), ˙γ(r)) = λr˙γ(r)
2
= g(w, a).
But we also have
g(w, a) = g(λar + u, a) = λra
2
= λr˙γ(0)
2
= λr˙γ(r)
2
.
Now we use the fact that
J(r) = (d exp
p
)
ra
w
and
˙γ(r) = (d exp
p
)
ra
a,
and we are done.
Corollary
(Local minimizing of length)
.
Let
a ∈ T
p
M
. We define
ϕ
(
t
) =
ta
,
and ψ(t) a piecewise C
1
curve in T
p
M for t ∈ [0, 1] such that
ψ(0) = 0, ψ(1) = a.
Then
length(exp
p
◦ψ) ≥ length(exp
p
◦ϕ) = a.
It is important to interpret this corollary precisely. It only applies to curves
with the same end point in
T
p
M
. If we have two curves in
T
p
M
whose end
points have the same image in
M
, then the result need not hold (the torus would
be a counterexample).
Proof.
We may of course assume that
ψ
never hits 0 again after
t
= 0. We write
ψ(t) = ρ(t)u(t),
where ρ(t) ≥ 0 and u(t) = 1. Then
ψ
0
= ρ
0
u + ρu
0
.
Then using the extended Gauss lemma, and the general fact that if
u
(
t
) is a unit
vector for all t, then u · u
0
=
1
2
(u · u)
0
= 0, we have
d
dx
(exp
p
◦ψ)(t)
2
=
(d exp
p
)
ψ(t)
ψ
0
(t)
2
= ρ
0
(t)
2
+ 2g(ρ
0
(t)u(t), ρ(t)u
0
(t)) + ρ(t)
2
(d exp
p
)
ψ(t)
u
0
(t)
2
= ρ
0
(t)
2
+ ρ(t)
2
(d exp
p
)
ψ(t)
u
0
(t)
2
,
Thus we have
length(exp
p
◦ψ) ≥
Z
1
0
ρ
0
(t) dt = ρ(1) − ρ(0) = a.
Notation. We write Ω(p, q) for the set of all piecewise C
1
curves from p to q.
We now wish to define a metric on M, in the sense of metric spaces.
Definition
(Distance)
.
Suppose
M
is connected, which is the same as it being
path connected. Let (p, q) ∈ M. We define
d(p, q) = inf
ξ∈Ω(p,q)
length(ξ),
where
To see this is indeed a metric, All axioms of a metric space are obvious, apart
from the nonnegativity part.
Theorem.
Let
p ∈ M
, and let
ε
be such that
exp
p

B(0,ε)
is a diffeomorphism
onto its image, and let U be the image. Then
–
For any
q ∈ U
, there is a unique geodesic
γ ∈
Ω(
p, q
) with
`
(
γ
)
< ε
.
Moreover,
`
(
γ
) =
d
(
p, q
), and is the unique curve that satisfies this property.
– For any point q ∈ M with d(p, q) < ε, we have q ∈ U .
–
If
q ∈ M
is any point,
γ ∈
Ω(
p, q
) has
`
(
γ
) =
d
(
p, q
)
< ε
, then
γ
is a
geodesic.
Proof.
Let
q
=
exp
p
(
a
). Then the path
γ
(
t
) =
exp
p
(
ta
) is a geodesic from
p
to
q
of length
a
=
r < ε
. This is clearly the only such geodesic, since
exp
p

B(0,ε)
is a diffeomorphism.
Given any other path ˜γ ∈ Ω(p, q), we want to show `(˜γ) > `(γ). We let
τ = sup
n
t ∈ [0, 1] : γ([0, t]) ⊆ exp
p
(B(0, r))
o
.
Note that if τ 6= 1, then we must have γ(τ) ∈ Σ
r
, the geodesic sphere of radius
r
, otherwise we can continue extending. On the other hand, if
τ
= 1, then we
certainly have
γ
(
τ
)
∈
Σ
r
, since
γ
(
τ
) =
q
. Then by local minimizing of length,
we have
`(˜γ) ≥ `(˜γ
[0,τ]
) ≥ r.
Note that we can always lift
˜γ[0, τ]
to a curve from 0 to
a
in
T
p
M
, since
exp
p
is
a diffeomorphism in B(0, ε).
By looking at the proof of the local minimizing of length, and using the same
notation, we know that we have equality iff τ = 1 and
ρ(t)
2
(d exp
p
)
ψ(t)
ψ(t)u
0
(t)
2
= 0
for all
t
. Since d
exp
p
is regular, this requires
u
0
(
t
) = 0 for all
t
(since
ρ
(
t
)
6
= 0
when
t 6
= 0, or else we can remove the loop to get a shorter curve). This implies
˜γ lifts to a straight line in T
p
M, i.e. is a geodesic.
Now given any
q ∈ M
with
r
=
d
(
p, q
)
< ε
, we pick
r
0
∈
[
r, ε
) and a path
γ ∈ Ω(p, q) such that `(γ) = r
0
. We again let
τ = sup
n
t ∈ [0, 1] : γ([0, t]) ⊆ exp
p
(B(0, r
0
))
o
.
If
τ 6
= 1, then we must have
γ
(
τ
)
∈
Σ
r
0
, but lifting to
T
p
M
, this contradicts the
local minimizing of length.
The last part is an immediate consequence of the previous two.
Corollary.
The distance
d
on a Riemannian manifold is a metric, and induces
the same topology on M as the C
∞
structure.
Definition
(Minimal geodesic)
.
A minimal geodesic is a curve
γ
: [0
,
1]
→ M
such that
d(γ(0), γ(1)) = `(γ).
One would certainly want a minimal geodesic to be an actual geodesic. This
is an easy consequence of what we’ve got so far, using the observation that a
subcurve of a minimizing geodesic is still minimizing.
Corollary.
Let
γ
: [0
,
1]
→ M
be a piecewise
C
1
minimal geodesic with constant
speed. Then γ is in fact a geodesic, and is in particular C
∞
.
Proof.
We wlog
γ
is unit speed. Let
t ∈
[0
,
1], and pick
ε >
0 such that
exp
p

B(0,ε)
is a diffeomorphism. Then by the theorem,
γ
[t,t+
1
2
ε]
is a geodesic.
So γ is C
∞
on (t, t +
1
2
ε), and satisfies the geodesic equations there.
Since we can pick
ε
continuously with respect to
t
by ODE theorems, any
t ∈ (0, 1) lies in one such neighbourhood. So γ is a geodesic.
While it is not true that geodesics are always minimal geodesics, this is locally
true:
Corollary. Let γ : [0, 1] ⊆ R → M be a C
2
curve with ˙γ constant. Then this
is a geodesic iff it is locally a minimal geodesic, i.e. for any
t ∈
[0
,
1), there exists
δ > 0 such that
d(γ(t), γ(t + δ)) = `(γ
[t,t+δ]
).
Proof.
This is just carefully applying the previous theorem without getting
confused.
To prove
⇒
, suppose
γ
is a geodesic, and
t ∈
[0
,
1). We wlog
γ
is unit speed.
Then pick
U
and
ε
as in the previous theorem, and pick
δ
=
1
2
ε
. Then
γ
[t,t+δ]
is a geodesic with length
< ε
between
γ
(
t
) and
γ
(
t
+
δ
), and hence must have
minimal length.
To prove the converse, we note that for each
t
, the hypothesis tells us
γ
[t,t+δ]
is a minimizing geodesic, and hence a geodesic, but the previous corollary. By
continuity,
γ
must satisfy the geodesic equation at
t
. Since
t
is arbitrary,
γ
is a
geodesic.
There is another sense in which geodesics are locally length minimizing.
Instead of chopping up a path, we can say it is minimal “locally” in the space
Ω(p, q). To do so, we need to give Ω(p, q) a topology, and we pick the topology
of uniform convergence.
Theorem.
Let
γ
(
t
) =
exp
p
(
ta
) be a geodesic, for
t ∈
[0
,
1]. Let
q
=
γ
(1).
Assume
ta
is a regular point for
exp
p
for all
t ∈
[0
,
1]. Then there exists
a neighbourhood of
γ
in Ω(
p, q
) such that for all
ψ
in this neighbourhood,
`(ψ) ≥ `(γ), with equality iff ψ = γ up to reparametrization.
Before we prove the result, we first look at why the two conditions are
necessary. To see the necessity of
ta
being regular, we can consider the sphere
and two antipodal points:
p
q
Then while the geodesic between them does minimize distance, it does not do so
strictly.
We also do not guarantee global minimization of length. For example, we
can consider the torus
T
n
= R
n
/Z
n
.
This has a flat metric from
R
n
, and the derivative of the exponential map is
the “identity” on
R
n
at all points. So the geodesics are the straight lines in
R
n
.
Now consider any two
p, q ∈ T
n
, then there are infinitely many geodesics joining
them, but typically, only one of them would be the shortest.
p
q
Proof.
The idea of the proof is that if
ψ
is any curve close to
γ
, then we can use
the regularity condition to lift the curve back up to
T
p
M
, and then apply our
previous result.
Write
ϕ
(
t
) =
ta ∈ T
p
M
. Then by the regularity assumption, for all
t ∈
[0
,
1],
we know
exp
p
is a diffeomorphism of some neighbourhood
W
(
t
) of
ϕ
(
t
) =
at ∈
T
p
M
onto the image. By compactness, we can cover [0
,
1] by finitely many such
covers, say W (t
1
), ··· , W (t
n
). We write W
i
= W (t
i
), and we wlog assume
0 = t
0
< t
1
< ··· < t
k
= 1.
By cutting things up, we may assume
γ([t
i
, t
i+1
]) ⊆ W
i
.
We let
U =
[
exp
p
(W
i
).
Again by compactness, there is some
ε <
0 such that for all
t ∈
[
t
i
, t
i+1
], we
have B(γ(t), ε) ⊆ W
i
.
Now consider any curve
ψ
of distance
ε
away from
γ
. Then
ψ
([
t
i
, t
i+1
])
⊆ W
i
.
So we can lift it up to
T
p
M
, and the end point of the lift is
a
. So we are done
by local minimization of length.
Note that the tricky part of doing the proof is to make sure the lift of
ψ
has
the same end point as
γ
in
T
p
M
, which is why we needed to do it neighbourhood
by neighbourhood.