8Connections
III Differential Geometry
8.2 Geodesics and parallel transport
One thing we can do with a connection is to define a geodesic as a path with
“no acceleration”.
Definition
(Geodesic)
.
Let
M
be a manifold with a linear connection
∇
. We
say that γ : I → M is a geodesic if
D
t
˙γ(t) = 0.
A natural question to ask is if geodesics exist. This is a local problem, so we
work in local coordinates. We try to come up with some ordinary differential
equations that uniquely specify a geodesic, and then existence and uniqueness
will be immediate. If we have a vector field
V ∈ J
(
γ
), we can write it locally as
V = V
j
∂
j
,
then we have
D
t
V =
˙
V
j
∂
j
+ V
j
∇
˙γ(t
0
)
∂
j
.
We now want to write this in terms of Christoffel symbols. We put
γ
=
(γ
1
, · · · , γ
n
). Then using the chain rule, we have
D
t
V =
˙
V
k
∂
k
+ V
j
˙γ
i
∇
∂
i
∂
j
= (
˙
V
k
+ V
j
˙γ
i
Γ
k
ij
)∂
k
.
Recall that γ is a geodesic if D
t
˙γ = 0 on I. This holds iff we locally have
¨γ
k
+ ˙γ
i
˙γ
j
Γ
k
ij
= 0.
As this is just a second-order ODE in
γ
, we get unique solutions locally given
initial conditions.
Theorem.
Let
∇
be a linear connection on
M
, and let
W ∈ T
p
M
. Then there
exists a geodesic γ : (−ε, ε) → M for some ε > 0 such that
˙γ(0) = W.
Any two such geodesics agree on their common domain.
More generally, we can talk about parallel vector fields.
Definition
(Parallel vector field)
.
Let
∇
be a linear connection on
M
, and
γ
:
I → M
be a path. We say a vector field
V ∈ J
(
γ
) along
γ
is parallel if
D
t
V (t) ≡ 0 for all t ∈ I.
What does this say? If we think of D
t
as just the usual derivative, this tells
us that the vector field
V
is “constant” along
γ
(of course it is not literally
constant, since each V (t) lives in a different vector space).
Example. A path γ is a geodesic iff ˙γ is parallel.
The important result is the following:
Lemma
(Parallel transport)
.
Let
t
0
∈ I
and
ξ ∈ T
γ(t
0
)
M
. Then there exists a
unique parallel vector field
V ∈ J
(
γ
) such that
V
(
t
0
) =
ξ
. We call
V
the parallel
transport of ξ along γ.
Proof.
Suppose first that
γ
(
I
)
⊆ U
for some coordinate chart
U
with coordinates
x
1
, · · · , x
n
. Then V ∈ J(γ) is parallel iff D
t
V = 0. We put
V =
X
V
j
(t)
∂
∂x
j
.
Then we need
˙
V
k
+ V
j
˙γ
i
Γ
k
ij
= 0.
This is a first-order linear ODE in
V
with initial condition given by
V
(
t
0
) =
ξ
,
which has a unique solution.
The general result then follows by patching, since by compactness, the image
of γ can be covered by finitely many charts.
Given this, we can define a map given by parallel transport:
Definition
(Parallel transport)
.
Let
γ
:
I → M
be a curve. For
t
0
, t
1
, we define
the parallel transport map
P
t
0
t
1
: T
γ(t
0
)
M → T
γ(t
1
)
M
given by ξ 7→ V
ξ
(t
1
).
It is easy to see that this is indeed a linear map, since the equations for
parallel transport are linear, and this has an inverse
P
t
1
t
0
given by the inverse
path. So the connection ∇ “connects” T
γ(t
0
)
M and T
γ(t
1
)
M.
Note that this connection depends on the curve
γ
chosen! This problem is in
general unfixable. Later, we will see that there is a special kind of connections
known as flat connections such that the parallel transport map only depends on
the homotopy class of the curve, which is an improvement.