2Vector fields
III Differential Geometry
2.2 Flows
What can we do with vector fields? In physics, we can imagine a manifold as all
of space, and perhaps a vector field specifies the velocity a particle should have
at that point. Now if you actually drop a particle into that space, the particle
will move according to the velocity specified. So the vector field generates a flow
of the particle. These trajectories are known as integral curves.
Definition
(Integral curve)
.
Let
X ∈ Vect
(
M
). An integral curve of
X
is a
smooth γ : I → M such that I is an open interval in R and
˙γ(t) = X
γ(t)
.
Example. Take M = R
2
, and let
X = x
∂
∂y
− y
∂
∂x
.
The field looks like this:
We would expect the integral curves to be circles. Indeed, suppose
γ
:
I → R
2
is
an integral curve. Write γ = (γ
1
, γ
2
). Then the definition requires
γ
0
1
(t)
∂
∂x
+ γ
0
2
(t)
∂
∂y
= γ
1
(t)
∂
∂y
− γ
2
(t)
∂
∂x
.
So the equation is
γ
0
1
(t) = −γ
2
(t)
γ
0
2
(t) = γ
1
(t).
For example, if our starting point is p = (1, 0), then we have
γ
1
(t) = cos t, γ
2
(t) = sin t.
We see that to find an integral curve, all we are doing is just solving ordinary
differential equations. We know that all ODEs have smooth and unique solutions,
and they have all the nice properties we can hope for. So we are going to get nice
corresponding results for integral solutions. However, sometimes funny things
happen.
Example. Take M = R, and
X = x
2
d
dx
.
Then if γ is an integral curve, it must satisfy:
γ
0
(t) = γ(t)
2
.
This means that the solution is of the form
γ(t) =
1
C − t
for C a constant. For example, if we want γ(0) =
1
2
, then we have
γ(t) =
1
2 − t
.
The solution to this ODE is defined only for
t <
2, so we can only have
I = (−∞, 2) at best.
We are going to prove that integral curves always exist. To do so, we need
to borrow some powerful theorems from ODE theory:
Theorem
(Fundamental theorem on ODEs)
.
Let
U ⊆ R
n
be open and
α
:
U →
R
n
smooth. Pick t
0
∈ R.
Consider the ODE
˙γ
i
(t) = α
i
(γ(t))
γ
i
(t
0
) = c
i
,
where c = (c
1
, · · · , c
n
) ∈ R
n
.
Then there exists an open interval
I
containing
t
0
and an open
U
0
⊆ U
such
that for every
c ∈ U
0
, there is a smooth solution
γ
c
:
I → U
satisfying the ODE.
Moreover, any two solutions agree on a common domain, and the function
Θ : I × U
0
→ U defined by Θ(t, c) = γ
c
(t) is smooth (in both variables).
Theorem
(Existence of integral curves)
.
Let
X ∈ Vect
(
M
) and
p ∈ M
. Then
there exists some open interval
I ⊆ R
with 0
∈ I
and an integral curve
γ
:
I → M
for X with γ(0) = p.
Moreover, if
˜γ
:
˜
I → M
is another integral curve for
X
, and
˜γ
(0) =
p
, then
˜γ = γ on I ∩
˜
I.
Proof.
Pick local coordinates for
M
centered at
p
in an open neighbourhood
U
.
So locally we write
X =
n
X
i=1
α
i
∂
∂x
i
,
where α
i
∈ C
∞
(U). We want to find γ = (γ
1
, · · · , γ
n
) : I → U such that
n
X
i=1
γ
0
i
(t)
∂
∂x
i
γ(t)
=
n
X
i=1
α
i
(γ(t))
∂
∂x
i
γ(t)
, γ
i
(0) = 0.
Since the
∂
∂x
i
form a basis, this is equivalent to saying
γ
i
(t) = α
i
(γ(t)), γ
i
(0) = 0
for all i and t ∈ I.
By the general theory of ordinary differential equations, there is an interval
I and a solution γ, and any two solutions agree on their common domain.
However, we need to do a bit more for uniqueness, since all we know is that
there is a unique integral curve lying in this particular chart. It might be that
there are integral curves that do wild things when they leave the chart.
So suppose
γ
:
I → M
and
˜γ
:
˜
I → M
are both integral curves passing
through the same point, i.e. γ(0) = ˜γ(0) = p.
We let
J = {t ∈ I ∩
˜
I : γ(t) = ˜γ(t)}.
This is non-empty since 0
∈ J
, and
J
is closed since
γ
and
˜γ
are continuous. To
show it is all of I ∩
˜
I, we only have to show it is open, since I ∩
˜
I is connected.
So let
t
0
∈ J
, and consider
q
=
γ
(
t
0
). Then
γ
and
˜γ
are integral curves of
X
passing through
q
. So by the first part, they agree on some neighbourhood of
t
0
.
So J is open. So done.
Definition
(Maximal integral curve)
.
Let
p ∈ M
, and
X ∈ Vect
(
M
). Let
I
p
be
the union of all I such that there is an integral curve γ : I → M with γ(0) = p.
Then there exists a unique integral curve
γ
:
I
p
→ M
, known as the maximal
integral curve.
Note that I
p
does depend on the point.
Example. Consider the vector field
X =
∂
∂x
on
R
2
\ {
0
}
. Then for any point
p
= (
x, y
), if
y 6
= 0, we have
I
p
=
R
, but if
y
= 0 and
x <
0, then
I
p
= (
−∞, −x
). Similarly, if
y
= 0 and
x >
0, then
I
p
= (−x, ∞).
Definition
(Complete vector field)
.
A vector field is complete if
I
p
=
R
for all
p ∈ M .
Given a complete vector field, we obtain a flow map as follows:
Theorem.
Let
M
be a manifold and
X
a complete vector field on
M
. Define
Θ
t
: R × M → M by
Θ
t
(p) = γ
p
(t),
where
γ
p
is the maximal integral curve of
X
through
p
with
γ
(0) =
p
. Then Θ
is a function smooth in p and t, and
Θ
0
= id, Θ
t
◦ Θ
s
= Θ
s+t
Proof.
This follows from uniqueness of integral curves and smooth dependence
on initial conditions of ODEs.
In particular, since Θ
t
◦ Θ
−t
= Θ
0
= id, we know
Θ
−1
t
= Θ
−t
.
So Θ
t
is a diffeomorphism.
More algebraically, if we write
Diff
(
M
) for the diffeomorphisms
M → M
,
then
R → Diff(M)
t 7→ Θ
t
is a homomorphism of groups. We call this a one-parameter subgroup of diffeo-
morphisms.
What happens when we relax the completeness assumption? Everything is
essentially the same whenever things are defined, but we have to take care of
the domains of definition.
Theorem. Let M be a manifold, and X ∈ Vect(M). Define
D = {(t, p) ∈ R × M : t ∈ I
p
}.
In other words, this is the set of all (t, p) such that γ
p
(t) exists. We set
Θ
t
(p) = Θ(t, p) = γ
p
(t)
for all (t, p) ∈ D. Then
(i) D is open and Θ : D → M is smooth
(ii) Θ(0, p) = p for all p ∈ M.
(iii)
If (
t, p
)
∈ D
and (
t,
Θ(
s, p
))
∈ D
, then (
s
+
t, p
)
∈ D
and Θ(
t,
Θ(
s, p
)) =
Θ(t + s, p).
(iv) For any t ∈ R, the set M
t
: {p ∈ M : (t, p) ∈ D} is open in M, and
Θ
t
: M
t
→ M
−t
is a diffeomorphism with inverse Θ
−t
.
This is really annoying. We now prove the following useful result that saves
us from worrying about these problems in nice cases:
Proposition.
Let
M
be a compact manifold. Then any
X ∈ Vect
(
M
) is
complete.
Proof. Recall that
D = {(t, p) : Θ
t
(p) is defined}
is open. So given
p ∈ M
, there is some open neighbourhood
U ⊆ M
of
p
and an
ε >
0 such that (
−ε, ε
)
× U ⊆ D
. By compactness, we can find finitely many
such U that cover M , and find a small ε such that (−ε, ε) × M ⊆ D.
In other words, we know Θ
t
(
p
) exists and
p ∈ M
and
|t| < ε
. Also, we
know Θ
t
◦
Θ
s
= Θ
t+s
whenever
|t|, |s| < ε
, and in particular Θ
t+s
is defined. So
Θ
Nt
= (Θ
t
)
N
is defined for all N and |t| < ε, so Θ
t
is defined for all t.