PMATH 465/665: Riemannian Geometry
Assignment 6; Solutions
[1] Let (
M, g
) be a Riemannian manifold of dimension
n
and let
p
∈
M
. Let
{
e
1
, . . . , e
n
}
be
any
orthonor
mal frame at
p
∈
M
.
For example, we can take the coordinate frame
{
∂
∂x
1
, . . . ,
∂
∂x
n
}
of a normal
coordinate chart centred at
p
. [Recall that in normal coordinates,
g
ij
(
p
) =
g
p
∂
∂x
i
p
,
∂
∂x
j
p
=
δ
ij
.]
Let
B
ε
(0) be an open subset of the domain of the restricted exponential map. Hence exp
p
:
B
ε
(0)
⊂
T
p
M
→
exp
p
(
B
ε
(0))
⊂
M
is a diffeomorphism. In fact, the map is described as follows. If
V
p
∈
B
ε
(0),
then exp
p
(
V
p
) is the point in
M
obtained by following the geodesic with initial point
p
and initial
unit velocity
1

V
p

gp
V
p
for a time
t
=

V
p

g
p
. In particular, all points in
U
= exp
p
(
B
ε
(0)) are connected
to
p
by a unique unitspeed geodesic starting at
p
.
We will define an orthonormal frame
{
E
1
, . . . , E
n
}
on
U
as follows. Fix
j
. Define for
q
∈
U
, define
E
j
(
q
) to be the parallel transport of
e
j
to
q
by the unique geodesic from
p
to
q
.
Since parallel
transport with respect to a metriccompatible connection preserves the inner product, this procedure
defines an orthonormal frame
{
E
1
, . . . , E
n
}
on
U
.
Each vector field
E
j
is smooth because of the
smooth dependence on initial conditions from the ODE theorem.
By construction, each
E
j
is parallel at
q
in the direction of the radial geodesics passing through
q
.
That is, if
γ
: [0
, ε
) is the unique unitspeed radial geodesic starting at
p
and passing through
q
at
time
t
∈
[0
, ε
), then
∇
γ
0
(
t
)
E
j
= 0. But for
t
= 0, we have
γ
(0) =
p
and every unit tangent vector
U
p
at
p
is the initial velocity of a unitspeed radial geodesic starting at
p
. Thus
∇
U
p
E
j
= 0, which
implies that (
∇
E
i
E
j
)(
p
) = 0, since
E
i

p
=
e
i
is a unit tangent vector at
p
.
[2] Let
∇
be a connection on the tangent bundle
TM
of a manifold
M
. As in the case of the LeviCivita
connection, we define the
curvature tensor
R
=
R
∇
of
∇
by
R
(
X, Y
)
Z
=
∇
X
(
∇
Y
Z
)
 ∇
Y
(
∇
X
Z
)
 ∇
[
X,Y
]
Z.
We say that the connection is
flat
if
R
∇
= 0. We proved in class that
R
= 0 if and only if there
exists a
parallel local frame
{
E
1
, . . . , E
n
}
near every point
p
∈
M
. [We also saw that the domain
U
of the parallel frame was topologically trivial – it was an open rectangle, hence connected and simply
connected.]
Let
∇
be a flat connection on
TM
. Then for any
p
∈
M
, there is a connected, simplyconnected, open
neighbourhood
U
of
p
in
M
, and a parallel frame
{
E
1
, . . . , E
n
}
on
U
. Let
V
p
=
V
i
E
i

p
∈
T
p
M
be a
tangent vector at
p
. Define a smooth vector field
V
on
U
by
V
=
V
i
E
i
. That is,
V
q
=
V
i
E
i

q
. The
vector field
V
is smooth because its component functions with respect to a smooth local frame are
smooth (in fact they are constant). Let
γ
be any path from
p
to
q
. Let
D
t
be covariant differentiation
along the curve
γ
. Then
(
D
t
V
)(
t
0
) =
dV
i
dt
t
=
t
0
E
i

γ
(
t
0
)
+
V
i
∇
γ
0
(
t
0
)
E
i
!