PMATH 465/665: Riemannian Geometry
Assignment 1; Solutions
[1] We want to show that every vector field
X
with compact support is complete.
That is, we want
to show that the integral curve of
X
through any
p
∈
M
is defined for all
t
∈
R
. We start with a
preliminary lemma.
Lemma.
Let
X
∈
Γ(
TM
), and let Θ be the flow of
X
. Suppose there exists an
ε >
0 such that,
for any point
p
∈
M
, the integral curve Θ
(
p
)
starting at
p
is defined for all
t
∈
(

ε, ε
). Then
X
is
complete.
Proof of lemma.
We’ll prove this by contradiction. Suppose there exists a point
p
∈
M
such that
the maximal domain
D
(
p
)
of definition of Θ
(
p
)
is (
a, b
) where either
a
6
=
∞
or
b
6
= +
∞
. The proof
is similar in both cases, so let us assume
b
6
= +
∞
.
Choose
t
0
∈
(
b

ε, b
), and let
q
= Θ
(
p
)
(
t
0
).
By hypothesis, the integral curve Θ
(
q
)
of
X
starting at
q
exists at least for
t
∈
(

ε, ε
).
Define
γ
: (

ε, t
0
+
ε
)
→
M
by
γ
(
t
) =
(
Θ
(
p
)
(
t
)
,

ε < t < b,
Θ
(
q
)
(
t

t
0
)
,
t
0

ε < t < t
0
+
ε.
This is welldefined, because in the overlap region (
b

ε, b
) because Θ
(
q
)
(
t

t
0
) = Θ
t

t
0
(
q
) =
Θ
t
◦
Θ

t
0
◦
Θ
t
0
(
p
) = Θ
t
(
p
) = Θ
(
p
)
(
t
) by the group law for Θ.
Hence
γ
is an integral curve of
X
starting at
p
, and is defined at least until
t
=
t
0
+
ε > b
, which is a contradiction.
We can now easily complete [no pun intended] the argument.
Let
K
= supp(
X
).
By the ODE
theorem, for each
p
∈
K
, there exists some
ε
p
>
0 and some open neighbourhood
U
p
of
p
such that
Θ
(
q
)
is defined at least on (

ε, ε
) for all
q
∈
U
p
. Since
K
is compact, the open cover
{
U
p
;
p
∈
K
}
of
K
has a finite subcover
{
U
p
1
, . . . , U
p
k
}
. Let
ε
= min
{
ε
1
, . . . , ε
k
}
, which is positive. Then Θ
(
p
)
(
t
) is
defined at least for
t
∈
(

ε, ε
) for all
p
∈
K
. If
p
∈
M
\
K
, then
p
is a singular point of
X
, so the
integral curve Θ
(
p
)
is defined for all
t
∈
R
at these points. Hence the hypotheses of the lemma are
satisfied, and we conclude that
X
is complete.
Corollary.
Let
M
be a compact manifold. Then any vector field
X
on
M
is complete.
Proof of corollary.
The set supp(
X
) is a closed subset of the compact Hausdorff space
M
, and
hence is compact.
[2] Let Θ be a flow on an
oriented
manifold. We want to show that for each
t
∈
R
, the diffeomorphism
Θ
t
is
orientationpreserving
wherever it is defined.
We will begin by deriving an explicit condition for a diffeomorphism
f
:
M
→
N
between two
oriented manifolds
M
and
N
to be orientationpreserving, that we will be able to verify in this
case. A diffeomorphism
f
is orientationpreserving if and only if its pushforward (
f
*
)
p
takes oriented
bases of
T
p
M
to oriented bases of
T
f
(
p
)
N
at every
p
∈
M
. If we consider only oriented coordinate
charts (
U, ϕ
) and (
V, ψ
) for
M
and
N
, then the coordinate frames
∂
∂x
1
p
, . . . ,
∂
∂x
n
p
of (
U, ϕ
) and
∂
∂y
1
q
, . . . ,
∂
∂y
n
q
are oriented bases of
T
p
M
and
T
q
N
, respectively, at every point in their domain.