3Geodesics
III Riemannian Geometry
3.4 Completeness and the Hopf–Rinow theorem
There are some natural questions we can ask about geodesics. For example, we
might want to know if geodesics can be extended to exist for all time. We might
also be interested if distances can always be realized by geodesics. It turns out
these questions have the same answer.
Definition
(Geodesically complete)
.
We say a manifold (
M, g
) is geodesically
complete if each geodesic extends for all time. In other words, for all
p ∈ M
,
exp
p
is defined on all of T
p
M.
Example. The upper half plane
H
2
= {(x, y) : y > 0}
under the induced Euclidean metric is not geodesically complete. However,
H
2
and R
2
are diffeomorphic but R
2
is geodesically complete.
The first theorem we will prove is the following:
Theorem.
Let (
M, g
) be geodesically complete. Then any two points can be
connected by a minimal geodesic.
In fact, we will prove something stronger — let
p ∈ M
, and suppose
exp
p
is
defined on all of
T
p
M
. Then for all
q ∈ M
, there is a minimal geodesic between
them.
To prove this, we need a lemma
Lemma. Let p, q ∈ M . Let
S
δ
= {x ∈ M : d(x, p) = δ}.
Then for all sufficiently small δ, there exists p
0
∈ S
δ
such that
d(p, p
0
) + d(p
0
, q) = d(p, q).
Proof.
For
δ >
0 small, we know
S
δ
= Σ
δ
is a geodesic sphere about
p
, and
is compact. Moreover,
d
(
·, q
) is a continuous function. So there exists some
p
0
∈ Σ
δ
that minimizes d( ·, q).
Consider an arbitrary
γ ∈
Ω(
p, q
). For the sake of sanity, we assume
δ <
d(p, q). Then there is some t such that γ(t) ∈ Σ
δ
, and
`(γ) ≥ d(p, γ(t)) + d(γ(t), q) ≥ d(p, p
0
) + d(p
0
, q).
So we know
d(p, q) ≥ d(p, p
0
) + d(p
0
, p).
The triangle inequality gives the opposite direction. So we must have equality.
We can now prove the theorem.
Proof of theorem.
We know
exp
p
is defined on
T
p
M
. Let
q ∈ M
. Let
q ∈ M
.
We want a minimal geodesic in Ω(
p, q
). By the first lemma, there is some
δ >
0
and p
0
such that
d(p, p
0
) = δ, d(p, p
0
) + d(p
0
, q) = d(p, q).
Also, there is some v ∈ T
p
M such that exp
p
v = p
0
. We let
γ
p
(t) = exp
p
t
v
v
.
We let
I = {t ∈ R : d(q, γ
p
(t)) + t = d(p, q)}.
Then we know
(i) δ ∈ I
(ii) I is closed by continuity.
Let
T = sup{I ∩ [0, d(p, q)]}.
Since
I
is closed, this is in fact a maximum. So
T ∈ I
. We claim that
T
=
d
(
p, q
).
If so, then γ
p
∈ Ω(p, q) is the desired minimal geodesic, and we are done.
Suppose this were not true. Then
T < d
(
p, q
). We apply the lemma to
˜p
=
γ
p
(
T
), and
q
remains as before. Then we can find
ε >
0 and some
p
1
∈ M
with the property that
d(p
1
, q) = d(γ
p
(T ), q) − d(γ
p
(T ), p
1
)
= d(γ
p
(T ), q) − ε
= d(p, q) − T − ε
Hence we have
d(p, p
1
) ≥ d(p, q) − d(q, p
1
) = T + ε.
Let γ
1
be the radial (hence minimal) geodesic from γ
p
(T ) to p
1
. Now we know
`(γ
p

[0,T ]
) + `(γ
1
) = T + ε.
So
γ
1
concatenated with
γ
p

[0,T ]
is a lengthminimizing geodesic from
p
to
p
1
,
and is hence a geodesic. So in fact
p
1
lies on
γ
p
, say
p
1
=
γ
p
(
T
+
s
) for some
s
. Then
T
+
s ∈ I
, which is a contradiction. So we must have
T
=
d
(
p, q
), and
hence
d(q, γ
p
(T )) + T = d(p, q),
hence d(q, γ
p
(T )) = 0, i.e. q = γ
p
(T ).
Corollary
(Hopf–Rinow theorem)
.
For a connected Riemannian manifold (
M, g
),
the following are equivalent:
(i) (M, g) is geodesically complete.
(ii) For all p ∈ M, exp
p
is defined on all T
p
M.
(iii) For some p ∈ M, exp
p
is defined on all T
p
M.
(iv) Every closed and bounded subset of (M, d) is compact.
(v) (M, d) is complete as a metric space.
Proof.
(i) and (ii) are equivalent by definition. (ii)
⇒
(iii) is clear, and we proved
(iii) ⇒ (i).
–
(iii)
⇒
(iv): Let
K ⊆ M
be closed and bounded. Then by boundedness,
K
is contained in
exp
p
(
B(0, R)
). Let
K
0
be the preimage of
K
under
exp
p
.
Then it is a closed and bounded subset of
R
n
, hence compact. Then
K
is
the continuous image of a compact set, hence compact.
– (iv) ⇒ (v): This is a general topological fact.
–
(v)
⇒
(i): Let
γ
(
t
) :
I → R
be a geodesic, where
I ⊆ R
. We wlog
˙γ ≡
1.
Suppose
I 6
=
R
. We wlog
sup I
=
a < ∞
. Then
lim
t→a
γ
(
t
) exist by
completeness, and hence
γ
(
a
) exists. Since geodesics are locally defined
near
a
, we can pick a geodesic in the direction of
lim
t→a
γ
0
(
t
). So we can
extend γ further, which is a contradiction.