PMATH 465/665: Riemannian Geometry
Assignment 2; Solutions
[1] Let
G
be a Lie group, and let
X
be a
leftinvariant
vector field on
G
.
[a] Let Θ be the flow of
X
. Recall from Assignment #1, problem #1, we proved a lemma that said: if
there exists an
ε >
0 such that,
for any point
g
∈
G
, the integral curve Θ
(
g
)
starting at
g
is defined
for all
t
∈
(

ε, ε
), then
X
is complete. Consider the point
p
=
e
, the identity element of
G
. There
exists some
ε >
0 such that Θ
(
e
)
is defined on (

ε, ε
). We claim that Θ
(
g
)
is defined on (

ε, ε
).
This would prove that
X
is complete. Define
γ
: (

ε, ε
)
→
G
by
γ
(
t
) =
g
·
Θ
(
e
)
(
t
) = (
L
g
◦
Θ
(
e
)
)(
t
),
where
·
is the group multiplication in
G
. We have
γ
(0) =
g
·
Θ
(
e
)
(0) =
g
·
e
=
g
. Furthermore,
using the fact that Θ
(
e
)
is an integral curve of
X
, and that
X
is leftinvariant, we find that
γ
0
(
t
0
) =
d
dt
t
=
t
0
γ
(
t
) =
d
dt
t
=
t
0
(
L
g
◦
Θ
(
e
)
)(
t
)
=
(
(
L
g
)
*
)
Θ
(
e
)
(
t
0
)
(
(Θ
(
e
)
)
0
(
t
0
)
)
=
(
(
L
g
)
*
)
Θ
(
e
)
(
t
0
)
(
X
Θ
(
e
)
(
t
0
)
)
=
X
g
·
Θ
(
e
)
(
t
0
)
=
X
γ
(
t
0
)
.
Thus
γ
is an integral curve of
X
and
γ
(0) =
g
, so
γ
= Θ
(
g
)
by uniqueness. Hence
X
is complete.
[b] All integral curves are smooth maps from their domain into the manifold.
So we just need to
prove the group homomorphism property. Since
X
is leftinvariant, its flow Θ is a global flow by
part [a]. Now for any
t, s
∈
R
, we have that
Θ
(
e
)
(
t
+
s
) = Θ
t
+
s
(
e
) = Θ
t
(Θ
s
(
e
))
= Θ
t
(Θ
(
e
)
(
s
)) = Θ
(Θ
(
e
)
(
s
))
(
t
)
.
In part [a] we showed that Θ
(
g
)
=
g
·
Θ
(
e
)
for any
g
∈
G
, so we have
Θ
(
e
)
(
t
+
s
) = Θ
(Θ
(
e
)
(
s
))
(
t
) = Θ
(
e
)
(
s
)
·
Θ
(
e
)
(
t
)
.
This says that Θ
(
e
)
is a group homomorphism.
[c] Let
γ
:
R
→
G
be a 1parameter subgroup of
G
. That is,
γ
is smooth and a group homomor
phism.
We want to show that
γ
is an integral curve of a leftinvariant vector field.
Since
γ
is
a homomorphism, we have
γ
(0) =
e
∈
G
. Let
X
e
=
γ
0
(0)
∈
T
e
G
, and define the leftinvariant
vector field
X
on
G
as usual by
X
g
=
(
(
L
g
)
*
)
e
(
X
e
)
∈
T
g
G
. We need to show that
γ
is an integral
curve of
X
.
Let
t
0
∈
R
.
By the chain rule, the fact that
γ
is a homomorphism, the fact that
γ
0
(0) =
X
e
, and the leftinvariance of
X
, we have
d
dt
t
=
t
0
γ
(
t
) =
d
ds
s
=0
γ
(
t
0
+
s
) =
d
ds
s
=0
γ
(
t
0
)
·
γ
(
s
)
=
d
ds
s
=0
(
L
γ
(
t
0
)
◦
γ
)(
s
) =
(
(
L
γ
(
t
0
)
)
*
)
γ
(0)
(
γ
0
(0))
=
(
(
L
γ
(
t
0
)
)
*
)
γ
(0)
(
X
e
) =
X
γ
(
t
0
)
·
e
=
X
γ
(
t
0
)
,
which is what we wanted to show.
[2] Let
X
be a vector field on
M
, and let Θ be the flow of
X
.
Then (Θ
t
)
*
:
T
*
Θ
t
(
p
)
M
→
T
*
p
M
is an
isomorphism whenever Θ
t
(
p
) is defined. Given
p
∈
M
and a (
k,
0) tensor
σ
on
M
, we define the
Lie
derivative
(
L
X
σ
)
p
to be the (
k,
0) tensor given by
(
L
X
σ
)
p
= lim
t
→
0
(Θ
t
)
*
(
σ
Θ
t
(
p
)
)

σ
p
t
=
d
dt
t
=0
(Θ
t
)
*
(
σ
Θ
t
(
p
)
)
.
1